$L\text{'}Hpital\text{'}s$ Rule can $be$ used $to$ evaluate the $limit,$ but $it$ must $be$ put $in$ the form $\frac{0}{0}or\frac{}{}.$ How

should the $limitbe$ rewritten?

$A.L=\underset{x}{lim}xln(1+\frac{a}{x})=\underset{x}{lim}\frac{ln(1+\frac{a}{x})}{x}$

$B.L=\underset{x}{lim}xln(1+\frac{a}{x})=\underset{x}{lim}\frac{x}{\frac{1}{ln(1\frac{+}{a}x)}}$

$C.L=\underset{x}{lim}xln(1+\frac{a}{x})=\underset{x}{lim}\frac{x}{ln(1+\frac{a}{x})}$

$D.L=\underset{x}{lim}xln(1+\frac{a}{x})=\underset{x}{lim}\frac{ln(1+\frac{a}{x})}{\frac{1}{x}}$

Apply $l\text{'}Hpital\text{'}s$ Rule $by$ finding the derivative $of$ the numerator and the denominator

separately.

$\underset{x}{lim}\frac{ln(1+\frac{a}{x})}{\frac{1}{x}}=\underset{x}{lim}$