40. L'hopital's Rule
Dated: 30-10-2024
Type \(\frac 0 0\)
We can have expressions like
\[\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\]
\[\lim_{x \to 0} \frac{\sin x}{x}\]
Which may or may not converge and what they do converge to is not immediately obvious.
Therefore, we need L'Hopital's Rule.
The Rule
If \(\lim f(x) = \lim g(x) = 0\) and \(\lim \frac{f^{\prime}(x)}{g^{\prime}(x)}\) gives us a finite value \(L\) or \(+\infty\) or \(-\infty\) then
\[\lim \frac{f(x)}{g(x)} = \lim \frac{f^{\prime}(x)}{g^{\prime}(x)}\]
Steps
- Check if \(\lim \frac{f(x)}{g(x)}\) is in
indeterminate form. If it is not then the rule cannot be used.
- Differentiate \(f(x)\) and \(g(x)\) separately.
- Find \(\lim \frac{f^{\prime}(x)}{g^{\prime}(x)}\) and if it is finite, \(+\infty\) or \(-\infty\) then
\[\lim \frac{f(x)}{g(x)} = \lim \frac{f^{\prime}(x)}{g^{\prime}(x)}\]
Example
\[\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\]
Step 1
\[\lim_{x \to 2} (x^2 - 4) = 0\]
\[\lim_{x \to 2} (x - 2) = 0\]
Step 2
\[\lim_{x \to 2}\frac{(x^{2}-4)}{x-2}\]
\[=\lim_{x \to 2}\frac{\frac{d}{dx}[x^{2}-4]}{\frac{d}{dx}[x-2]} \]
\[=\lim_{x \to 2}\frac{2x}{1}=4\]
Step 3
\[\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4\]
Type \(\frac \infty \infty\)
If \(\lim f(x) = \lim g(x) = \infty\) and \(\lim \frac{f^{\prime}(x)}{g^{\prime}(x)}\) gives us a finite value \(L\) or \(+\infty\) or \(-\infty\) then
\[\lim \frac{f(x)}{g(x)} = \lim \frac{f^{\prime}(x)}{g^{\prime}(x)}\]
Example
Evaluate
\[\lim_{x \to +\infty} = \frac {x}{e^x}\]
Step 1
\[\lim_{x \to +\infty} x = +\infty\]
\[\lim_{x \to +\infty} e^x = +\infty\]
Step 2
\[\lim_{x\rightarrow \infty}\frac{x}{e^{x}}\]
\[=\lim_{x\rightarrow \infty}\frac{\frac{d}{dx}[x]}{\frac{d}{dx}[e^{x}]}\]
\[=\lim_{x\rightarrow \infty}\frac{1}{e^{x}}=0\]
Step 3
\[\lim_{x \to +\infty} = \frac {x}{e^x} = 0\]
Type \(0 \cdot \infty\)
If we have \(f(x) = 0\) and \(g(x) = \infty\) then a product limit \(\lim f(x) \cdot g(x)\) is of the form \(0 \cdot \infty\) and to apply the rule on it, we need to convert it into either \(\frac 0 0\) form or \(\frac \infty \infty\) form.
Example
Evaluate
\[\lim_{x \to 0^+} x \ln x\]
Step 1
\[\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac 1 x}\]
\[\lim_{x \to 0^+} \ln x = -\infty\]
\[\lim_{x \to 0^+} \frac 1 x = + \infty\]
Step 2
\[\lim_{x \to 0^+} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} \frac{1}{x}}\]
\[= \lim_{x \to 0^+} \frac{1/x}{- 1/x^2} = \lim_{x \to 0^+} (-x) = 0\]
Step 3
\[\lim_{x \to 0^+} x \ln x = 0\]
Type \(0^0\), \(\infty^0\), \(1^\infty\), \(\infty^\infty\)
These usually come of as the result of
\[\lim_{x \to a} y = \lim_{x \to a}f(x)^{g(x)}\]
To solve these,
\[\lim_{x\rightarrow a}\ln y=\lim_{x\rightarrow a}\left(\ln (f(x))^{g(x)}\right)\]
\[=\lim_{x\rightarrow a}\left(g(x)\ln f(x)\right)\]
If we know \(\lim_{x\rightarrow a}\ln y\) then determining the other side becomes easier.
Example
Show that
\[\lim_{x\rightarrow0}(1+x)^{1/x}=e\]
Step 1
\[\lim_{x \to 0}(1 + x) = 1\]
\[\lim_{x \to 0} \frac 1 x = \infty\]
Step 2
\[y=(1+x)^{1/x}\]
\[\ln y=\ln(1+x)^{1/x}\]
\[=\frac{1}{x}\ln(1+x)\]
\[=\frac{\ln(1+x)}{x}\]
\[\lim_{x\rightarrow0}\ln y=\lim_{x\rightarrow0}\frac{\ln(1+x)}{x}\]
\[\lim_{x \to 0} \ln (1 + x) = \ln (1) = 0\]
\[\lim_{x \to 0} x = 0\]
Step 3
\[\lim_{x\rightarrow0}\ln y=\lim_{x\rightarrow0}\frac{\ln(1+x)}{x}\]
\[=\lim_{x\rightarrow0}\frac{\frac{d}{dx}\ln(1+x)}{\frac{d}{dx}x}\]
\[ = \lim_{x \to 0}\frac{1 / (1 + x)}{1} = 1\]
This shows that as \(x \to 0\), \(\ln y \to 0\).
\[\because e^{\ln y} \to e^1\]
\[\therefore x \to 0 \implies y \to e\]
\[\lim_{x\rightarrow0}(1+x)^{1/x}=e\]
References
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