Question

Solve $\underset{x\to 4}{lim}\frac{(x^3+64)\ln (x-3)}{x^2-16}$

• A. $-8$
• B. $8$
• C. $16$
• D. $-16$

Answer 1

The correct option is C $16$

$\underset{x\to 4}{lim}\frac{(x^3+64)\left(\ln (x-3)\right)}{(x^2-16)}=l$ (say) [is of the form $\left(\frac{0}{0}\right)$]

$l=\underset{x\to 4}{lim}(x^3+64).\underset{x\to 4}{lim}\frac{\ln (x-3)}{x^2-16}\left[\underset{x\to a}{lim}f\left(x\right).g\left(x\right)=\underset{x\to a}{lim}f\left(x\right).\underset{x\to a}{lim}g\left(x\right)\right]$

$l=2\left(64\right).\underset{x\to 4}{lim}\frac{\frac{1}{x-3}}{2x}$ [L' Hospital Rule]

$l=\frac{2\left(64\right)}{2\left(4\right)}=16$