L’Hopital’s rule for other forms

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We use derivatives to give us a “short-cut” for computing limits.

To deal with indeterminate forms, we have L’Hôpital’s rule.

L’Hôpital’s Rule Let $f(x)$ and $g(x)$ be functions that are differentiable near $a$ . If $\lim _{x \to a} f(x) = \lim _{x \to a}g(x) = 0 \qquad \text {or} \pm \infty ,$ and $\lim _{x \to a} \frac {f'(x)}{g'(x)}$ exists, and $g'(x) \neq 0$ for all $x$ near $a$ , then $\lim _{x \to a} \frac {f(x)}{g(x)} = \lim _{x \to a} \frac {f'(x)}{g'(x)}.$

For the other indeterminate forms, L’Hôpital’s Rule does not apply. Our approach will be to modify the form so we can apply L’Hôpital’s Rule.

Indeterminate forms involving multiplication

$\zeroTimesInfty$ -forms arise from a limit of the form: $\lim _{x\to a} f(x)g(x)$ . To write $f(x)g(x)$ as a fraction, we remember $f(x)g(x) = \frac {f(x)}{\frac {1}{g(x)}} = \frac {g(x)}{\frac {1}{f(x)}}$

Let’s work through an example.

Compute $\lim _{x\to 0^+} x\ln x.$

This doesn’t appear to be suitable for L’Hôpital’s Rule. As $x$ approaches zero, $\ln x$ goes to $-\infty$ , so the product looks like $(\text {something very small})\cdot (\text {something very large and negative}).$ This product could be anything. A careful analysis is required. Write $x\ln x = \frac {\ln x}{\frac {1}{x}} = \frac {\ln x}{x^{-1}}.$ Set $f(x) = \ln (x)$ and $g(x) = x^{-1}$ . Since both functions are differentiable near zero and $\lim _{x\to 0+} \ln (x) = -\infty \qquad \text {and}\qquad \lim _{x\to 0+} x^{-1} = \infty ,$ we may apply L’Hôpital’s rule. Write with me $f'(x) = \answer [given]{x^{-1}}$ and $g'(x) = \answer [given]{-x^{-2}},$ so $\begin{align*} \lim_{x\to 0^+} x\ln x &= \lim_{x\to 0^+} \frac{\ln x}{x^{-1}} \\ &= \lim_{x\to 0^+} \frac{x^{-1}}{-x^{-2}}\\ &=\lim_{x\to 0^+} -x \\ &= 0. \end{align*}$

One way to interpret this is that since $\lim _{x\to 0^+}x\ln x = 0$ , the function $x$ approaches zero much faster than $\ln x$ approaches $-\infty$ .

Indeterminate forms involving subtraction

There are two basic cases here, we’ll do an example of each.

Compute $\lim _{x\to 0} \left (\cot (x) - \csc (x)\right ).$

Here we simply need to write each term as a fraction, $\begin{align*} \lim_{x\to 0} \left(\cot(x) - \csc(x)\right) &= \lim_{x\to 0} \left(\frac{\cos(x)}{\sin(x)} - \frac{1}{\sin(x)}\right)\\ &= \lim_{x\to 0} \frac{\cos(x)-1}{\sin(x)} \end{align*}$

Setting $f(x) = \cos (x)-1$ and $g(x)=\sin (x)$ , both functions are differentiable near zero and $\lim _{x\to 0}(\cos (x)-1)=\lim _{x\to 0}\sin (x) = 0.$ We may now apply L’Hôpital’s rule. Write with me $f'(x) = \answer [given]{-\sin (x)}$ and $g'(x) = \answer [given]{\cos (x)},$ so

$\begin{align*} \lim_{x\to 0} \left(\cot(x) - \csc(x)\right) &= \lim_{x\to 0} \frac{\cos(x)-1}{\sin(x)}\\ &= \lim_{x\to 0} \frac{-\sin(x)}{\cos(x)} \\ &=0. \end{align*}$

Sometimes one must be slightly more clever.

Compute $\lim _{x\to \infty }\left (\sqrt {x^2+x}-x\right ).$

Again, this doesn’t appear to be suitable for L’Hôpital’s Rule. A bit of algebraic manipulation will help. Write with me $\begin{align*} \lim_{x\to\infty}\left(\sqrt{x^2+x}-x\right) &= \lim_{x\to\infty}\left(x\left(\sqrt{1+1/x}-1\right)\right)\\ &=\lim_{x\to\infty}\frac{\sqrt{1+1/x}-1}{x^{-1}} \end{align*}$

Now set $f(x) = \sqrt {1+1/x}-1$ , $g(x) = x^{-1}$ . Since both functions are differentiable for large values of $x$ and $\lim _{x\to \infty } (\sqrt {1+1/x}-1) = \lim _{x\to \infty }x^{-1} = 0,$ we may apply L’Hôpital’s rule. Write with me $f'(x) = \answer [given]{(1/2)(1+1/x)^{-1/2}\cdot (-x^{-2})}$ and $g'(x) = \answer [given]{-x^{-2}}$ so

$\begin{align*} \lim_{x\to\infty}\left(\sqrt{x^2+x}-x\right) &= \lim_{x\to\infty}\frac{\sqrt{1+1/x}-1}{x^{-1}} \\ &= \lim_{x\to\infty}\frac{(1/2)(1+1/x)^{-1/2}\cdot(-x^{-2})}{-x^{-2}} \\ &= \lim_{x\to\infty} \frac{1}{2\sqrt{1+1/x}}\\ &= \frac{1}{2}. \end{align*}$

Exponential Indeterminate Forms

There is a standard trick for dealing with the indeterminate forms $\text {\oneToInfty },\quad \text {\zeroToZero },\quad \text {\inftyToZero }.$ Given $u(x)$ and $v(x)$ such that $\lim _{x\to a}u(x)^{v(x)}$ falls into one of the categories described above, rewrite as $\lim _{x\to a}e^{v(x)\ln (u(x))}$ and then examine the limit of the exponent $\lim _{x\to a} v(x)\ln (u(x)) = \lim _{x\to a} \frac {\ln (u(x))}{v(x)^{-1}}$ using L’Hôpital’s rule. Since these forms are all very similar, we will only give a single example.

Compute $\lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x.$

Write $\lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x = \lim _{x\to \infty }e^{x\ln \left (1 + \frac {1}{x}\right )}.$ So now look at the limit of the exponent $\lim _{x\to \infty } x\ln \left (1 + \frac {1}{x}\right ) = \lim _{x\to \infty } \frac {\ln \left (1 + \frac {1}{x}\right )}{x^{-1}}.$ Setting $f(x) = \ln \left (1 + \frac {1}{x}\right )$ and $g(x) = x^{-1}$ , both functions are differentiable for large values of $x$ and $\lim _{x\to \infty }\ln \left (1 + \frac {1}{x}\right )=\lim _{x\to \infty }x^{-1} = 0.$ We may now apply L’Hôpital’s rule. Write $f'(x) = \answer [given]{\frac {-x^{-2}}{1 + \frac {1}{x}}}$ and $g'(x) = \answer [given]{-x^{-2}},$ so $\begin{align*} \lim_{x\to\infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{x^{-1}} &= \lim_{x\to\infty} \frac{\frac{-x^{-2}}{1 + \frac{1}{x}}}{-x^{-2}} \\ &=\lim_{x\to\infty} \frac{1}{1 + \frac{1}{x}}\\ &=1. \end{align*}$

Hence, $\lim _{x\to \infty }\left (1 + \frac {1}{x}\right )^x = \lim _{x\to \infty }e^{x\ln \left (1 + \frac {1}{x}\right )} =e^{1} = e.$