  ### Home > APCALC > Chapter 6 > Lesson 6.5.1 > Problem6-168

6-168.

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

1. $\lim\limits _ { x \rightarrow 1 } \frac { x ^ { 2 } - 1 } { \operatorname { ln } ( x ) }$

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

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

$2$

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.

$\lim \limits_{x\to 0^{+}}\frac{e^{x}}{\ln(x)}\rightarrow \frac{1}{-\infty }=0$

What does $\lim \limits_{x\to 0^–}\frac{e^{x}}{\ln(x)}=\underline{\ \ \ \ \ \ \ \ }?$

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 $f '(x)$, or use l'Hôpital'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'Hôpital's Rule, you will need to convert the argument into a single fraction that

$\rightarrow \frac{0}{0} \text{ or }\frac{\infty }{\infty }.$

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

To evaluate this limit, you will have to use l'Hôpital'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 $\tan(x)=\frac{\sin(x)}{\cos(x)}$.

1. $\lim\limits _ { x \rightarrow 0 } \frac { x ^ { 2 } } { \operatorname { sin } ^ { 2 } ( x ) }$

See the hint in part (a).