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Home/Calculus/Chapter 4/L'Hospital & Taylor

CALC-4.4

5-6 hours

L'Hospital's Rule and Taylor's Theorem

Master powerful techniques for evaluating limits and approximating functions: L'Hospital's rule for indeterminate forms and Taylor's theorem for polynomial approximations.

Learning Objectives

Apply L'Hospital's rule for 0/0 and ∞/∞ indeterminate forms

Convert other indeterminate forms to apply L'Hospital's rule

Understand Taylor's theorem with Peano remainder

Apply Taylor's theorem with Lagrange remainder for error bounds

Use Cauchy's remainder form in special cases

Compute Taylor expansions of standard functions

Use Taylor expansions to evaluate limits

Apply Taylor expansions to approximate function values

1. L'Hospital's Rule

Theorem 4.13: L'Hospital's Rule (0/0 form)

Suppose $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ , and $g'(x) \neq 0$ near $a$ . If $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists (or is $\pm\infty$ ), then:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

Theorem 4.14: L'Hospital's Rule (∞/∞ form)

The same conclusion holds when $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = \pm\infty$ .

Example 4.13: Basic Application

Evaluate $\lim_{x \to 0} \frac{e^x - 1}{x}$ .

This is $\frac{0}{0}$ form. Apply L'Hospital's rule:

\lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1

Example 4.14: Converting Other Forms

Evaluate $\lim_{x \to 0^+} x \ln x$ ( $0 \cdot (-\infty)$ form).

Rewrite as $\frac{\ln x}{1/x}$ ( $\frac{-\infty}{\infty}$ form):

\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0

Proof of Theorem 4.13 (L'Hospital's Rule - 0/0 case):

Using Cauchy's Mean Value Theorem:

For $x \neq a$ , there exists $\xi$ between $a$ and $x$ such that:

\frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(\xi)}{g'(\xi)}

Since $f(a) = g(a) = 0$ :

\frac{f(x)}{g(x)} = \frac{f'(\xi)}{g'(\xi)}

As $x \to a$ , $\xi \to a$ , so if $\lim_{x \to a}\frac{f'(x)}{g'(x)} = L$ exists, the limit equals $L$ .

∎

Remark 4.1: Indeterminate Forms

L'Hospital's rule applies directly to:

\frac{0}{0}

\frac{\infty}{\infty}

0 \cdot \infty

(convert)

\infty - \infty

(convert)

For exponential forms $0^0$ , $1^\infty$ , $\infty^0$ , take logarithms first.

Example 4.14b: Repeated Application

Problem: Evaluate $\lim_{x \to 0} \frac{e^x - 1 - x - \frac{x^2}{2}}{x^3}$ .

Solution:

This is $\frac{0}{0}$ . Apply L'Hospital's three times:

\lim \frac{e^x - 1 - x}{3x^2} = \lim \frac{e^x - 1}{6x} = \lim \frac{e^x}{6} = \frac{1}{6}

Example 4.14c: ∞ - ∞ Form

Problem: Evaluate $\lim_{x \to 0} \left(\frac{1}{x} - \frac{1}{\sin x}\right)$ .

Solution:

Combine fractions: $\frac{\sin x - x}{x \sin x}$ (now $\frac{0}{0}$ )

\lim \frac{\sin x - x}{x \sin x} = \lim \frac{\cos x - 1}{\sin x + x\cos x} = \lim \frac{-\sin x}{2\cos x - x\sin x} = \frac{0}{2} = 0

Example 4.14d: 1^∞ Form

Problem: Evaluate $\lim_{x \to 0} (1 + x)^{1/x}$ .

Solution:

Let $y = (1 + x)^{1/x}$ . Then $\ln y = \frac{\ln(1 + x)}{x}$ .

\lim_{x \to 0} \frac{\ln(1 + x)}{x} = \lim_{x \to 0} \frac{1/(1+x)}{1} = 1

Therefore $\lim y = e^1 = e$ .

Example 4.14e: 0^0 Form

Problem: Evaluate $\lim_{x \to 0^+} x^x$ .

Solution:

Let $y = x^x$ . Then $\ln y = x \ln x$ .

We showed $\lim_{x \to 0^+} x \ln x = 0$ earlier.

Therefore $\lim y = e^0 = 1$ .

Remark 4.1b: When L'Hospital's Rule Fails

Be careful! L'Hospital's rule can fail when:

✗The limit of derivatives doesn't exist (but the original limit might!)
✗Applying it leads to more complex expressions
✗It's applied to a non-indeterminate form

Classic Failure Example:

\lim_{x \to \infty} \frac{x}{\sqrt{x^2 + 1}} = \lim_{x \to \infty} \frac{1}{\frac{x}{\sqrt{x^2+1}}} \text{ (circular!)}

The correct answer is 1 (divide by $x$ ).

Example 4.14f: L'Hospital at Infinity

Problem: Evaluate $\lim_{x \to \infty} \frac{\ln x}{x}$ .

Solution:

This is $\frac{\infty}{\infty}$ . Apply L'Hospital's:

\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0

This shows: polynomials grow faster than logarithms.

Example 4.14g: Exponential vs Polynomial

Problem: Evaluate $\lim_{x \to \infty} \frac{x^n}{e^x}$ for any $n \in \mathbb{N}$ .

Solution:

Apply L'Hospital's $n$ times:

\lim \frac{x^n}{e^x} = \lim \frac{nx^{n-1}}{e^x} = \cdots = \lim \frac{n!}{e^x} = 0

This shows: exponentials grow faster than any polynomial.

Example 4.14h: ∞^0 Form

Problem: Evaluate $\lim_{x \to \infty} x^{1/x}$ .

Solution:

Let $y = x^{1/x}$ . Then $\ln y = \frac{\ln x}{x}$ .

\lim_{x \to \infty} \frac{\ln x}{x} = \lim_{x \to \infty} \frac{1/x}{1} = 0

Therefore $\lim y = e^0 = 1$ .

Remark 4.1c: Growth Rate Hierarchy

As $x \to \infty$ , growth rates from slowest to fastest:

\ln x \ll x^\alpha \ll e^x \ll e^{x^2} \ll x^x

Here $f \ll g$ means $\lim f/g = 0$ .

Example 4.14i: Composite Function Limit

Problem: Evaluate $\lim_{x \to 0^+} (\sin x)^{\tan x}$ .

Solution:

This is $0^0$ form. Let $y = (\sin x)^{\tan x}$ .

\ln y = \tan x \cdot \ln(\sin x) = \frac{\ln(\sin x)}{\cot x}

As $x \to 0^+$ : $\frac{-\infty}{+\infty}$ . Apply L'Hospital:

\lim \frac{\cos x / \sin x}{-\csc^2 x} = \lim \frac{\cos x \sin x}{-1} = 0

Therefore $\lim y = e^0 = 1$ .

Example 4.14j: A Tricky Limit

Problem: Evaluate $\lim_{x \to 0} \frac{e^x - e^{\sin x}}{x - \sin x}$ .

Solution:

Both numerator and denominator $\to 0$ as $x \to 0$ .

\lim \frac{e^x - e^{\sin x}}{x - \sin x} = \lim \frac{e^x - e^{\sin x} \cos x}{1 - \cos x}

Still $\frac{0}{0}$ . Apply again:

= \lim \frac{e^x - e^{\sin x}(\cos^2 x - \sin x)}{\sin x}

At $x = 0$ : numerator $= 1 - 1 = 0$ . Continue or use Taylor:

Using Taylor: $e^x - e^{\sin x} = e^x(1 - e^{\sin x - x}) \approx -(\sin x - x) = \frac{x^3}{6} + o(x^3)$

And $x - \sin x = \frac{x^3}{6} + o(x^3)$ . So limit = 1.

2. Taylor's Theorem

Theorem 4.15: Taylor's Theorem (Peano Remainder)

If $f$ has $n$ derivatives at $x_0$ , then:

f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k + o((x-x_0)^n)

Theorem 4.16: Taylor's Theorem (Lagrange Remainder)

If $f \in C^n[a,b] \cap D^{n+1}(a,b)$ , then for any $x, x_0 \in [a,b]$ , there exists $\xi$ between $x$ and $x_0$ :

R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}

Theorem 4.17: Taylor's Theorem (Cauchy Remainder)

Under the same conditions, there exists $\theta \in (0, 1)$ :

R_n(x) = \frac{f^{(n+1)}(x_0 + \theta(x-x_0))}{n!}(1-\theta)^n(x-x_0)^{n+1}

Proof of Theorem 4.15 (Taylor's Theorem - Peano Form):

We prove by induction. For $n = 0$ :

f(x) = f(x_0) + o(1) \quad \text{(by continuity)}

Assume true for $n-1$ . Define $R_n(x) = f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$ .

We need to show $R_n(x) = o((x-x_0)^n)$ .

Apply L'Hospital's rule:

\lim_{x \to x_0} \frac{R_n(x)}{(x-x_0)^n} = \lim_{x \to x_0} \frac{R_n'(x)}{n(x-x_0)^{n-1}}

Since $R_n'(x)$ is the $(n-1)$ -th Taylor remainder of $f'$ , by induction, this limit is 0.

∎

Remark 4.2: Comparing Remainder Forms

Peano Remainder

R_n = o((x-x_0)^n)

Best for limits; local behavior only

Lagrange Remainder

R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}h^{n+1}

Best for error bounds

Cauchy Remainder

R_n = \frac{f^{(n+1)}(x_0+\theta h)}{n!}(1-\theta)^n h^{n+1}

Useful for special cases

Definition 4.5: Taylor Polynomial

The n-th Taylor polynomial of $f$ centered at $x_0$ is:

T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k

When $x_0 = 0$ , this is called the Maclaurin polynomial.

Example 4.15b: Computing Taylor Polynomial

Problem: Find the 4th-degree Taylor polynomial of $f(x) = e^x$ at $x_0 = 0$ .

Solution:

For $e^x$ , all derivatives equal $e^x$ , so $f^{(k)}(0) = 1$ for all $k$ .

T_4(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}

Example 4.15c: Taylor Polynomial for sin x

Problem: Find the 5th-degree Maclaurin polynomial of $\sin x$ .

Solution:

Derivatives at 0: $\sin 0 = 0$ , $\cos 0 = 1$ , $-\sin 0 = 0$ , $-\cos 0 = -1$ , ...

T_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} = x - \frac{x^3}{6} + \frac{x^5}{120}

Remark 4.2b: Why Taylor Expansions Work

Taylor's theorem tells us that smooth functions can be locally approximated by polynomials.

Key insight: The polynomial $T_n(x)$ matches $f(x)$ at $x_0$ in:

• Value: $T_n(x_0) = f(x_0)$
• First derivative: $T_n'(x_0) = f'(x_0)$
• All derivatives up to order n

Example 4.15d: Error Bound Example

Problem: Approximate $\sin(0.1)$ using the 3rd-degree Taylor polynomial and bound the error.

Solution:

$T_3(x) = x - \frac{x^3}{6}$ , so $T_3(0.1) = 0.1 - \frac{0.001}{6} \approx 0.09983$ .

Lagrange remainder: $|R_3| = \left|\frac{f^{(4)}(\xi)}{4!}(0.1)^4\right| = \left|\frac{\sin\xi}{24}(0.0001)\right|$

Since $|\sin\xi| \leq 1$ : $|R_3| \leq \frac{0.0001}{24} \approx 4.2 \times 10^{-6}$ .

Corollary 4.5: Uniqueness of Taylor Polynomial

If $P_n(x)$ is any polynomial of degree $\leq n$ such that $f(x) - P_n(x) = o((x-x_0)^n)$ , then $P_n = T_n$ .

Example 4.15e: Taylor with Lagrange Error

Problem: Approximate $\cos(0.1)$ using $T_2$ and bound the error.

Solution:

$T_2(x) = 1 - \frac{x^2}{2}$ for $\cos x$ .

T_2(0.1) = 1 - \frac{0.01}{2} = 0.995

Lagrange remainder with $f^{(3)}(x) = \sin x$ :

|R_2| = \left|\frac{\sin \xi}{3!}(0.1)^3\right| \leq \frac{0.001}{6} \approx 1.67 \times 10^{-4}

Actual: $\cos(0.1) \approx 0.995004...$

Remark 4.2c: Choosing Between Remainder Forms

Use Peano When:

• Evaluating limits as $x \to x_0$
• You only care about local behavior
• Precise error bound not needed

Use Lagrange When:

• You need an error bound
• Computing numerical approximations
• Proving inequalities

Example 4.15f: Higher-Order Taylor

Problem: Find $T_4$ for $f(x) = \ln(1 + x)$ and estimate the error at $x = 0.5$ .

Solution:

T_4(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}

At $x = 0.5$ :

T_4(0.5) = 0.5 - 0.125 + 0.0417 - 0.0156 \approx 0.4011

Error bound using $|f^{(5)}(\xi)| = \frac{4!}{(1+\xi)^5} \leq 24$ :

|R_4| \leq \frac{24}{5!}(0.5)^5 = \frac{24 \cdot 0.03125}{120} = 0.00625

Actual: $\ln(1.5) \approx 0.4055$ . True error $\approx 0.0044$ .

Proof of Theorem 4.16 (Lagrange Remainder):

Define $R_n(x) = f(x) - T_n(x)$ and consider the auxiliary function:

\phi(t) = f(x) - T_n(x) - \frac{R_n(x)}{(x-x_0)^{n+1}}(t-x_0)^{n+1}

$\phi(x_0) = 0$ and $\phi(x) = 0$ .

By Rolle's theorem, $\exists \xi$ between $x_0$ and $x$ : $\phi'(\xi) = 0$ .

Computing $\phi'(t)$ and evaluating at $t = \xi$ gives:

R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}

∎

Definition 4.6: Big-O and Little-o Notation

$f(x) = O(g(x))$ as $x \to a$ means $|f(x)| \leq M|g(x)|$ for some constant $M$ near $a$ .

$f(x) = o(g(x))$ as $x \to a$ means $\lim_{x \to a} \frac{f(x)}{g(x)} = 0$ .

Example: $\sin x = x + O(x^3)$ means $\sin x - x$ is bounded by a constant times $x^3$ .

3. Standard Taylor Expansions

Maclaurin Series (x₀ = 0)

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + o(x^n)

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots + (-1)^n\frac{x^{2n+1}}{(2n+1)!} + o(x^{2n+2})

\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^n\frac{x^{2n}}{(2n)!} + o(x^{2n+1})

\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots + (-1)^{n-1}\frac{x^n}{n} + o(x^n)

(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \cdots + \binom{\alpha}{n}x^n + o(x^n)

Example 4.15: Using Taylor for Limits

Evaluate $\lim_{x \to 0} \frac{x - \sin x}{x^3}$ .

Using $\sin x = x - \frac{x^3}{6} + o(x^3)$ :

\frac{x - \sin x}{x^3} = \frac{x - (x - \frac{x^3}{6} + o(x^3))}{x^3} = \frac{\frac{x^3}{6} + o(x^3)}{x^3} \to \frac{1}{6}

Example 4.16: Approximating e

Estimate $e$ with error less than $10^{-5}$ .

Using $e = e^1 = \sum_{k=0}^n \frac{1}{k!} + R_n$ where $|R_n| < \frac{e}{(n+1)!} < \frac{3}{(n+1)!}$ .

For $n = 8$ : $\frac{3}{9!} \approx 8.3 \times 10^{-6} < 10^{-5}$ .

e \approx \sum_{k=0}^{8} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \cdots + \frac{1}{40320} \approx 2.71828

Remark 4.3: Deriving New Expansions from Known Ones

You can derive new Taylor expansions by:

Substitution: Replace $x$ with $x^2$ , $-x$ , etc.
Differentiation: Differentiate term by term
Integration: Integrate term by term
Multiplication: Multiply two series

Example 4.16b: Expansion by Substitution

Problem: Find the Maclaurin expansion of $e^{-x^2}$ .

Solution:

Start with $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$

Substitute $u = -x^2$ :

e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!}

Example 4.16c: Expansion by Differentiation

Problem: Find the expansion of $\frac{1}{(1-x)^2}$ .

Solution:

Start with $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$

Differentiate both sides:

\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + \cdots = \sum_{n=0}^{\infty} (n+1)x^n

Example 4.16d: Expansion by Integration

Problem: Find the expansion of $\arctan x$ .

Solution:

Note that $\frac{d}{dx}\arctan x = \frac{1}{1+x^2}$ .

Expand $\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots$

Integrate term by term:

\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}

Example 4.16e: Limit Evaluation with Taylor

Problem: Evaluate $\lim_{x \to 0} \frac{\cos x - 1 + \frac{x^2}{2}}{x^4}$ .

Solution:

Using $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots$ :

\cos x - 1 + \frac{x^2}{2} = \frac{x^4}{24} - \frac{x^6}{720} + \cdots

\frac{\cos x - 1 + \frac{x^2}{2}}{x^4} = \frac{1}{24} - \frac{x^2}{720} + \cdots \to \frac{1}{24}

Remark 4.3b: Key Standard Expansions

Memorize these expansions:

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}

\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}

\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}

\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}

\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n

(1+x)^\alpha = \sum_{n=0}^{\infty} \binom{\alpha}{n} x^n

Example 4.16f: Hyperbolic Functions

Problem: Find the Maclaurin expansions of $\sinh x$ and $\cosh x$ .

Solution:

Using $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$ :

\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}

\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}

Note: These are like $\sin x$ and $\cos x$ but without alternating signs!

Example 4.16g: Expansion of tan x

Problem: Find the first few terms of the Maclaurin series for $\tan x$ .

Solution:

Using $\tan x = \frac{\sin x}{\cos x}$ with long division or derivatives:

\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \cdots

The coefficients involve Bernoulli numbers. Converges for $|x| < \frac{\pi}{2}$ .

Remark 4.3c: Radius of Convergence

Infinite Radius:

$e^x, \sin x, \cos x, \sinh x, \cosh x$

Radius = 1:

$\ln(1+x), \frac{1}{1-x}, (1+x)^\alpha$

Example 4.16h: Combining Expansions

Problem: Expand $e^{\sin x}$ up to $x^4$ .

Solution:

Let $u = \sin x = x - \frac{x^3}{6} + O(x^5)$ .

Then $e^u = 1 + u + \frac{u^2}{2} + \frac{u^3}{6} + \cdots$

e^{\sin x} = 1 + \left(x - \frac{x^3}{6}\right) + \frac{1}{2}\left(x - \frac{x^3}{6}\right)^2 + \frac{1}{6}x^3 + \cdots

= 1 + x + \frac{x^2}{2} + \frac{x^3}{6} - \frac{x^3}{6} + O(x^4) = 1 + x + \frac{x^2}{2} + O(x^4)

Corollary 4.7: Even and Odd Functions

Even functions ( $f(-x) = f(x)$ ) have only even powers in their Maclaurin series.

Odd functions ( $f(-x) = -f(x)$ ) have only odd powers.

Examples: $\cos x$ is even; $\sin x$ is odd.

4. Applications

Example 4.17: Approximating sqrt(2)

Problem: Use Taylor expansion to approximate $\sqrt{2}$ .

Solution:

Write $\sqrt{2} = \sqrt{1 + 1} = (1 + 1)^{1/2}$ .

Using $(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2}x^2 + \cdots$ with $x = 1$ , $\alpha = 1/2$ :

\sqrt{2} \approx 1 + \frac{1}{2} - \frac{1}{8} + \frac{1}{16} - \cdots

Note: This converges slowly. Better to expand around $x = 1$ using $f(x) = \sqrt{1+x}$ .

Example 4.17b: Finding Series Coefficients

Problem: Find the coefficient of $x^5$ in the expansion of $e^x \sin x$ .

Solution:

Multiply the series:

e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \cdots

\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots

Coefficient of $x^5$ :

1 \cdot \frac{1}{120} + 1 \cdot 0 + \frac{1}{2} \cdot (-\frac{1}{6}) + \frac{1}{6} \cdot 0 + \frac{1}{24} \cdot 1 = \frac{1}{120} - \frac{1}{12} + \frac{1}{24} = -\frac{1}{30}

Remark 4.4: Taylor vs L'Hospital

For limit problems, Taylor expansion is often faster and cleaner than L'Hospital's rule:

Use Taylor when:

• Multiple applications of L'Hospital would be needed
• You need the exact order of the limit
• The function involves compositions

Use L'Hospital when:

• The function is simple
• One or two applications suffice
• Taylor expansion is not obvious

Example 4.17c: Taylor Expansion at Non-Zero Point

Problem: Find the Taylor expansion of $\ln x$ around $x_0 = 1$ .

Solution:

$f(x) = \ln x$ , $f'(x) = \frac{1}{x}$ , $f''(x) = -\frac{1}{x^2}$ , $f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n}$

At $x_0 = 1$ : $f^{(n)}(1) = (-1)^{n-1}(n-1)!$

\ln x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-1)^n}{n}

Example 4.17d: Physical Application: Small Angle Approximation

Application: For small angles, $\sin\theta \approx \theta$ (in radians).

From Taylor: $\sin\theta = \theta - \frac{\theta^3}{6} + \frac{\theta^5}{120} - \cdots$

For $\theta = 0.1$ rad (5.7°):

$\sin(0.1) = 0.09983...$
First approximation: $0.1$ (error < 0.02%)
With correction: $0.1 - 0.001/6 \approx 0.09983$

This approximation is fundamental in physics for pendulums, optics, etc.

Example 4.17e: Computing π Using Taylor Series

Application: Use $\arctan(1) = \frac{\pi}{4}$ to compute $\pi$ .

From the expansion $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ :

\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots

Note: This converges very slowly! Better formulas exist (Machin's formula).

Example 4.17f: Complex Limit Using Taylor

Problem: Evaluate $\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}$ .

Solution:

Using Taylor expansions:

\sin x = x - \frac{x^3}{6} + o(x^3)

\tan x = x + \frac{x^3}{3} + o(x^3)

Therefore:

\tan x - \sin x = \frac{x^3}{3} + \frac{x^3}{6} + o(x^3) = \frac{x^3}{2} + o(x^3)

\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2}

Remark 4.5: Taylor Expansion Pitfalls

Radius of convergence: $\ln(1+x)$ only valid for $-1 < x \leq 1$
Alternating series: Be careful with signs in $\cos x$ , $\ln(1+x)$
Center point: Expansion around $x_0 = 0$ may not converge at your target

Example 4.17g: Numerical Integration Preview

Application: Approximate $\int_0^1 e^{-x^2} dx$ .

Using $e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots$ :

\int_0^1 e^{-x^2} dx \approx \int_0^1 \left(1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}\right) dx

= \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42}\right]_0^1 = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} \approx 0.743

Actual value: $\approx 0.7468$ . Error < 0.5%.

Example 4.17h: Asymptotic Expansion

Problem: Find the behavior of $\sqrt{1+x} - \sqrt{1-x}$ as $x \to 0$ .

Solution:

Using $(1+u)^{1/2} = 1 + \frac{u}{2} - \frac{u^2}{8} + \cdots$ :

\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + O(x^3)

\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} + O(x^3)

\sqrt{1+x} - \sqrt{1-x} = x + O(x^3)

The difference is approximately $x$ for small $x$ .

Corollary 4.6: Taylor Expansion of Compositions

If $f$ and $g$ have Taylor expansions and $g(x_0) = 0$ , then:

f(g(x)) = f(0) + f'(0)g(x) + \frac{f''(0)}{2}g(x)^2 + \cdots

Substitute the expansion of $g$ and collect terms by degree.

Example 4.17i: Proving Euler's Formula Preview

Insight: Taylor expansions lead to Euler's formula $e^{ix} = \cos x + i\sin x$ .

Expanding $e^{ix}$ and grouping real/imaginary parts:

e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \cdots

= \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right) + i\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right)

= \cos x + i\sin x

Example 4.17j: Approximating Definite Integrals

Problem: Estimate $\int_0^{0.5} \frac{\sin x}{x} dx$ .

Solution:

Using $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \cdots$ :

\int_0^{0.5} \frac{\sin x}{x} dx \approx \int_0^{0.5} \left(1 - \frac{x^2}{6} + \frac{x^4}{120}\right) dx

= \left[x - \frac{x^3}{18} + \frac{x^5}{600}\right]_0^{0.5} = 0.5 - 0.00694 + 0.00005 \approx 0.4931

Remark 4.6: Comparison: L'Hospital vs Taylor

When to use each method:

Situation	Best Method
Simple 0/0 or ∞/∞	L'Hospital (1-2 applications)
Higher-order limits	Taylor (more efficient)
Numerical approximation	Taylor with error bound
Proving inequalities	Taylor with Lagrange remainder

Example 4.17k: Final Challenge Problem

Problem: Evaluate $\lim_{x \to 0} \frac{e^x - e^{-x} - 2\sin x}{x - \sin x}$ .

Solution:

Expand each term:

e^x - e^{-x} = 2x + \frac{2x^3}{3!} + O(x^5) = 2x + \frac{x^3}{3} + O(x^5)

2\sin x = 2x - \frac{2x^3}{6} + O(x^5) = 2x - \frac{x^3}{3} + O(x^5)

x - \sin x = \frac{x^3}{6} + O(x^5)

Numerator: $\frac{x^3}{3} + \frac{x^3}{3} + O(x^5) = \frac{2x^3}{3} + O(x^5)$

\lim_{x \to 0} \frac{\frac{2x^3}{3}}{\frac{x^3}{6}} = \frac{2/3}{1/6} = 4

Historical Context

L'Hospital's Rule (1696): Named after Guillaume de l'Hôpital, though actually discovered by Johann Bernoulli.

Taylor's Theorem (1715): Brook Taylor published this result, though earlier forms were known to Gregory and Newton.

Maclaurin Series (1742): Colin Maclaurin popularized the special case of Taylor series centered at 0.

L'Hospital & Taylor Practice

Questions

Correct

Accuracy

Evaluate

\lim_{x \to 0} \frac{\sin x}{x}

using L'Hospital's rule.

Easy

Not attempted

Evaluate

\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}

Medium

Not attempted

What is the Maclaurin series of

e^x

up to

x^3

Easy

Not attempted

The Lagrange remainder for Taylor's theorem is:

Medium

Not attempted

Evaluate

\lim_{x \to 0} \frac{x - \sin x}{x^3}

using Taylor expansion.

Medium

Not attempted

For which form can L'Hospital's rule NOT be directly applied?

Easy

Not attempted

Evaluate

\lim_{x \to 0^+} x \ln x

Medium

Not attempted

Use Taylor's theorem to estimate

e

with error less than

10^{-5}

Hard

Not attempted

Evaluate

\lim_{x \to 0} \frac{1 - \cos x}{x^2}

Easy

Not attempted

What is the coefficient of

x^4

in the Maclaurin series of

\cos x

Easy

Not attempted

Evaluate

\lim_{x \to 0} \frac{\tan x - x}{x^3}

Medium

Not attempted

What type of indeterminate form is

\lim_{x \to \infty} x^{1/x}

Medium

Not attempted

Using Taylor expansion,

e^x - 1 - x \approx

for small

x

Medium

Not attempted

The Lagrange remainder bounds the error by:

Medium

Not attempted

Frequently Asked Questions

When does L'Hospital's rule fail?

When the conditions aren't met (not 0/0 or ∞/∞), when the limit of derivatives doesn't exist, or when applying it creates a more complex form.

What's the difference between Peano and Lagrange remainders?

Peano remainder R_n = o((x-x₀)ⁿ) only describes behavior as x → x₀. Lagrange remainder gives an explicit formula with ξ, useful for error bounds.

How do I know which Taylor expansion to use?

Use the expansion centered at the point where you're evaluating (usually x₀ = 0 for Maclaurin). Keep enough terms so the remainder is negligible.

Can I use Taylor expansion instead of L'Hospital's rule?

Yes! Often Taylor expansion is faster and more insightful. For limits like (sin x - x)/x³, one Taylor expansion gives the answer directly.

How many terms should I keep in a Taylor expansion?

Keep enough terms so the highest power in the numerator matches or exceeds the power in the denominator. For limits, stop when you can determine the leading behavior.

What's the binomial expansion for non-integer exponents?

(1+x)^α = Σ C(α,n)x^n where C(α,n) = α(α-1)...(α-n+1)/n!. This converges for |x| < 1.

Can Taylor series diverge?

Yes! The series may only converge in a limited radius. For example, ln(1+x) only converges for -1 < x ≤ 1.

How do I handle limits at infinity with Taylor?

Substitute u = 1/x and let u → 0. Then apply standard Maclaurin expansions.

What if I need to evaluate a limit but don't know the Taylor expansion?

You can always compute Taylor coefficients directly: f^(n)(x₀)/n!. Or try L'Hospital's rule as an alternative.

Why is the Maclaurin series of sin x only odd powers?

Because sin is an odd function: sin(-x) = -sin(x). Odd functions have only odd powers in their Taylor series centered at 0.

Key Takeaways

L'Hospital's Rule

\lim \frac{f}{g} = \lim \frac{f'}{g'}

For 0/0 or ∞/∞ forms

Taylor's Theorem

f(x) = \sum \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k + R_n

Lagrange Remainder

R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}

Key Expansions

$e^x, \sin x, \cos x, \ln(1+x), (1+x)^\alpha$

Study Tips

L'Hospital Strategy

First check if it's 0/0 or ∞/∞. If not, convert using algebra or logarithms. Apply the rule, but verify the new limit exists.

Taylor for Limits

For complex limits, Taylor is often faster. Expand to the order of the denominator, then simplify and take the limit.

Memorize Standard Series

Know $e^x$ , $\sin x$ , $\cos x$ , $\ln(1+x)$ , and $(1+x)^\alpha$ by heart. Derive others from these.

Error Bounds

Use Lagrange remainder for numerical estimates. Bound $|f^{(n+1)}(\xi)|$ over the interval to get the maximum error.

Common Mistakes to Avoid

✗
Applying L'Hospital without checking form
Only use for 0/0 or ∞/∞. Other forms need conversion first.
✗
Keeping too few Taylor terms
Make sure numerator order matches or exceeds denominator order.
✗
Circular application of L'Hospital
Sometimes applying the rule leads back to the original limit.
✗
Confusing Peano and Lagrange remainders
Use Peano for limits, Lagrange for error bounds.

What's Next?

In the next module, you will learn about Applications of Derivatives: using derivatives to analyze functions, find extrema, and solve optimization problems.

Monotonicity and first derivative test
Concavity and second derivative test
Curve sketching techniques
Optimization problems

Chapter Summary

This module covered powerful techniques for evaluating limits and approximating functions:

1.L'Hospital's Rule handles 0/0 and ∞/∞ indeterminate forms via derivatives.
2.Other Forms like 0·∞, ∞−∞, 0⁰, 1^∞, ∞⁰ require conversion first.
3.Taylor's Theorem provides polynomial approximations with explicit error bounds.
4.Standard Maclaurin Series for common functions are essential tools.