Chapter 1.8, Problem 83E

Indeterminate Forms and I’Hopital’s Rule,

Let f and g be differentiable over an open interval containing $x=a$

If

$\underset{x\to a}{\lim }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{0}{0}or\underset{x\to a}{\lim }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\pm \infty }{\pm \infty }$

and if $\underset{x\to a}{\lim }\left(\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}\right)$

exists, then

$\underset{x\to a}{\lim }\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\underset{x\to a}{\lim }\left(\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}\right)$.

The forms $0\text{/}0$ and $\pm \infty /\pm \infty$ are said to be indeterminate. In such cases, the limit may exist, and I’Hopital’s Rule a way to find the limit using differentiation. For example, in Example 1 of section 1.1, we showed that

$\underset{x\to 1}{\lim }\left(\frac{x^2-1}{x-1}\right)=2$.

Since, for $x=1$ we have $\left(x^2-1\right)\left(x-1\right)=0/0$, we differentiate the numerator and denominator separately. And reevaluate the limit:

$\underset{x\to 1}{\lim }\left(\frac{x^2-1}{x-1}\right)=\underset{x\to 1}{\lim }\left(\frac{2x}{1}\right)=2$.

Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate from.

$\underset{x\to \infty }{\lim }\left(\frac{4x^2+x-3}{2x^2+1}\right)$