Question

The value of $\underset{x\to 3}{lim}\frac{\sqrt{x+6}-\sin (x-3)-3}{(x-3)\cos (x-3)}$ is equal to

• A. $\frac{5}{6}$
• B. $-\frac{5}{6}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$
  

Answer 1

Answer: (B)

$-\frac{5}{6}$

${lim}_{x\to 3}\frac{\sqrt{x+6}-\sin (x-3)-3}{(x-3)\cos (x-3)}=l$ (say)

This is of the form $\frac{0}{0}$, apply L' Hospital Rule

$l={lim}_{x\to 3}\frac{\frac{1}{2}\left(\frac{1}{\sqrt{x+6}}\right)-\cos (x-3)-0}{(x-3)(-\sin (x-3))+\left(\cos (x-3)\right)\left(1\right)}$

$l={lim}_{x\to 3}\frac{\frac{1}{2}\left(\frac{1}{\sqrt{x+6}}\right)-\cos (x-3)}{-(x-3)\sin (x-3)+\cos (x-3)}$

$l=\frac{\frac{1}{2}\left(\frac{1}{\sqrt{3+6}}\right)-1}{0+1}$

$l=\frac{\frac{1}{6}-1}{1}$

$l=\frac{-5}{6}$