### Home > CALC > Chapter 6 > Lesson 6.5.1 > Problem6-169

6-169.

Find the exact value of each of the following limits. Note that finding the exact value requires that the limit be found analytically rather than numerically.

1. $\lim\limits_ { x \rightarrow 1 } \frac { x ^ { 2 } - 1 } { \operatorname { ln } x }$

$\rightarrow \frac{0}{0},\text{ use l'Hopital's Rule }$

$=\lim\limits_{x\rightarrow 1}\frac{2x}{\frac{1}{x}}$

$2$

1. $\lim\limits_ { x \rightarrow 0 } \frac { e ^ { x }} { \operatorname { ln } x }$

Careful! Be sure to check:
For a limit to exist, the limit from the left must agree with the limit from the right.
We know that $\lim\limits_{x\rightarrow 0^{+}}\frac{e^{x}}{\text{ln}x}\rightarrow \frac{1}{-\infty }=0,$
What does $\lim\limits_{x\rightarrow 0^–}\frac{e^{x}}{\text{ln}x}=\underline{\ \ \ \ \ \ \ \ }?$

1. $\lim\limits_ { x \rightarrow 2 } \frac { 2 ^ { x } - 4 } { \operatorname { sin } ^ { - 1 } ( x - 2 ) }$

$\rightarrow \frac{0}{0}$

1. $\lim\limits_ { h \rightarrow 0 } \frac { \operatorname { ln } ( x + h ) - \operatorname { ln } x } { h }$

Recognize that this is Hana's Definition of the Derivative, and find $f^\prime(x)$, or use l'Hopital's Rule.

1. $\lim\limits_ { x \rightarrow 0 } ( \frac { 1 } { x } - \frac { 1 } { \operatorname { sin } x } )$

$→(0-0)$, which is also an indeterminate form.
Before using l'Hopital's Rule, you will need to convert the argument into a single fraction that
$\rightarrow \frac{0}{0} \text{ or }\frac{\infty }{\infty }.$

$=\lim\limits_{x\rightarrow 0}\left ( \frac{1}{x}-\frac{1}{\text{sin}x} \right )=\lim\limits_{x\rightarrow 0}\frac{\text{sin}x-x}{x\text{sin}x}\rightarrow \frac{0}{0}$

To solve this limit, you will have to use l'Hopital's Rule twice.

1. $\lim\limits_ { x \rightarrow \pi / 2 } ( \operatorname { tan } x \cdot \operatorname { ln } ( \operatorname { sin } x ) )$

To convert this limit into $\frac{0}{0}$ form, let $\operatorname{tan}x=\frac{\text{sin}x}{\text{cos}x}$, then write the argument as a single fraction.

1. $\lim\limits_ { x \rightarrow 0 } \frac { x ^ { 2 } } { \operatorname { sin } ^ { 2 } x }$

See hint in part (a).