We cannot apply L'Hospital's Rule directly for the indeterminate form \( \infty - \infty \).
Usually for problems of this form we must manipulate the expression before applying L'Hospital's Rule.
The most common techniques are:
• Taking reciprocals (convert terms like \( a - b \) into a quotient)
• Taking LCMs and simplifying the combined expression
After these algebraic steps, the expression becomes either \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), where L'Hospital's Rule can be applied safely.
PROBLEM 9
Evaluate \( \displaystyle \lim_{x\to \frac{\pi}{2}} \big( \sec x - \tan x \big) \)
Given
\[ \lim_{x\to\frac{\pi}{2}} (\sec x - \tan x) \quad (\infty - \infty) \]
\[ = \lim_{x\to\frac{\pi}{2}} \left( \frac{1}{\cos x} - \frac{\sin x}{\cos x} \right) \]
\[ = \lim_{x\to\frac{\pi}{2}} \frac{1 - \sin x}{\cos x} \qquad (0/0) \]
(Applying L'Hospital's Rule) \[ = \lim_{x\to\frac{\pi}{2}} \frac{0 - \cos x}{-\sin x} \]
\[ = \frac{0}{-1} \]
\[ = 0 \]
PROBLEM 10
Evaluate \( \displaystyle \lim_{x\to0} \left( \frac{1}{x} - \frac{1}{e^{x}-1} \right) \)
Given
\[ \lim_{x\to0} \left( \frac{1}{x} - \frac{1}{e^{x}-1} \right) \quad \left( \infty - \infty \right) \]
\[ = \lim_{x\to0} \frac{e^{x}-1 - x}{x(e^{x}-1)} \quad \left( \frac{0}{0} \right) \]
(Applying L'Hospital's Rule) \[ = \lim_{x\to0} \frac{e^{x}-1}{x e^{x} + (e^{x}-1)\cdot 1} \]
\[ = \lim_{x\to0} \frac{e^{x}-1}{x e^{x} + e^{x} - 1} \quad \left( \frac{0}{0} \right) \]
(Applying L'Hospital's Rule again) \[ = \lim_{x\to0} \frac{e^{x}}{x e^{x} + e^{x} + e^{x}} \]
\[ = \frac{1}{0 + 1 + 1} = \frac{1}{2} \]
\[ = \frac{1}{2} \]
EASIER WAY TO DO PROBLEM 10
\[ \lim_{x\to0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) \]
Given \[ \lim_{x\to0}\left(\frac{1}{x}-\frac{1}{e^{x}-1}\right) \;(\infty-\infty) \]
\[ = \lim_{x\to0}\frac{e^{x}-1-x}{x(e^{x}-1)} \;\bigg(\frac{0}{0}\bigg) \]
Now we multiply and divide by \(x\). Watch how the extra \(x^{2}\) and \(x\) appear:
\[ = \lim_{x\to0} \frac{e^{x}-1-x}{x(e^{x}-1)} \] \[ = \lim_{x\to0} \frac{e^{x}-1-x}{x} \cdot \frac{1}{e^{x}-1} \] \[ = \lim_{x\to0} \frac{e^{x}-1-x}{x^{2}} \cdot \frac{x}{e^{x}-1} \]
\[ =\left( \lim_{x\to0}\frac{e^{x}-1-x}{x^{2}} \right) \left( \lim_{x\to0}\frac{x}{e^{x}-1} \right) \]
\[ =\lim_{x\to0}\frac{e^{x}-1-x}{x^{2}} \;\bigg(\frac{0}{0}\bigg) \]
Apply L'Hospital's Rule once:
\[ = \lim_{x\to0}\frac{e^{x}-1}{2x} \;\bigg(\frac{0}{0}\bigg) \]
\[ = \frac12 \lim_{x\to0}\frac{e^{x}-1}{x} \]
\[ = \frac12 \cdot 1 \]
\[ = \frac12 \]
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