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Resolving indeterminate expressions

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We encounter indeterminate expressions when we compute some limits, always when this limit is the derivative.

The two main kinds come from fractions: $$\frac{\infty}{\infty}, \quad \frac{0}{0} \gets .$$

L'Hopital's Rule. $$\lim_{\substack{x \to a \\ \text{or} \\ x \to \infty}} \frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$$ if the latter exists, provided $$\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = 0 $$ or $$\lim_{x\to a} f(x) = \lim_{x\to a} g(x) = \infty$$

Warning: This is not the Quotient Rule!

ExampleCompute $$\lim_{x \to \infty} \frac{x^{2} - 3}{2x^{2} - x + 1}$$ The old method was to divide all like this $\dfrac{/x^{2}}{/x^{2}}$. Instead apply L'Hopital's Rule $$=\lim_{x \to \infty} \frac{2x}{4x-1}$$ This is still indeterminate! So apply LR again. $$=\lim_{x \to \infty} \frac{2}{4} = \frac{1}{2}.$$

Note: Before you apply L'Hopital's Rule verify that this is indeed an indeterminate expression. Otherwise you get $$\lim_{x\to 1}\frac{x^{2}}{x} = \lim_{x \to 1}\frac{2x}{1} = 2. ???$$

The old method only works for rational functions...

Example. Compute $$\lim_{x\to 1} \frac{\ln x}{x-1} \rightarrow \frac{0}{0}!$$ $$=\lim_{x \to 1} \frac{\frac{1}{x}}{1} = \lim_{x \to 1} \frac{1}{x} = 1.$$

Example. Compute $$\lim_{x\to \infty}\frac{e^{x}}{x^{2}} \qquad \rightarrow \frac{\infty}{\infty} $$ LR $$=\lim_{x\to\infty} \frac{e^{x}}{2x} \qquad \rightarrow \frac{\infty}{\infty}$$ LR $$=\lim_{x \to \infty} \frac{e^{x}}{2} = \infty. $$ Similarly $$\lim_{x \to \infty} \frac{e^{x}}{x^{1000}} = \infty. $$


What about other indeterminate expressions?

Possible types correspond to algebraic operations: product, difference, power.

Idea: Convert to fraction.

Product: $0\cdot\infty$ type form

$$\lim_{x\to 0^{+}} x \ln x$$ How do we apply LR here?

Convert to a fraction by dividing by the reciprocal. $$\begin{aligned} = \lim_{x\to 0^{+}} \frac{\ln x}{\frac{1}{x}} \qquad \to \frac{-\infty}{\infty} \\ &= \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^{2}}} = \lim_{x\to 0^{+}} -x = 0 \end{aligned}$$

Difference $\infty - \infty$ type form

$$\lim_{x\to \infty}(\sqrt{x^{2} + x} - x)$$ Convert to a fraction by "multiplying" by the conjugate $$\begin{aligned} =\lim_{x\to \infty} \frac{(\sqrt{x^{2} + x} - x)(\sqrt{x^{2} + x} + x)}{(\sqrt{x^{2} + x} + x)} &= \lim_{x\to\infty} \frac{x^{2} + x - x^{2}}{\sqrt{x^{2} + x} + x} \\ &= \lim_{x\to \infty}\frac{x}{\sqrt{x^{2} + x} + x} \qquad \to \frac{\infty}{\infty} \\ \text{LR} &= \lim_{x\to\infty} \frac{1}{\frac{1}{2}\left(x^{2} + x\right)^{-\frac{1}{2}}\cdot\left(2x+1\right) + 1} \\ & = \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x}}+1} = \frac{1}{2} \end{aligned}$$

LR failed here are we failed to simplify the denominator. Instead by divide by $x$ as before.

Try $\sec x \cdot \tan x$ at $(\frac{\pi}{2})$.

Power $0^{0}$, $\infty^{0}$, $1^{\infty}$ type form

$$\lim_{x \to \infty}x^{\frac{1}{x}} \qquad \to \infty^{0}$$ How do we convert to a fraction?

Idea: Use log! $$\ln x^{\frac{1}{x}} = \frac{1}{x}\ln x. $$ Then $$\begin{aligned} \ln(\lim_{x \to \infty} x^{\frac{1}{x}}) & \overset{?}{=} \lim_{x \to \infty}\left(\ln x^{\frac{1}{x}}\right) \\ &= \lim_{x \to \infty} \left(\frac{1}{x}\ln x\right) \\ &= \lim_{x\to\infty}\frac{\ln x}{x} \qquad \to \frac{-\infty}{\infty} \\ \text{LR} &= \lim_{x\to\infty}\frac{\frac{1}{x}}{1} = 0 \end{aligned}$$ So $$\lim_{x\to\infty} x^{\frac{1}{x}} = e^{0} = 1. $$ What property of limits we used at the end?