Question

If $f\left(x\right)=\begin{array}{c}1-cos3x,x\ne 0\\ k,x=0\end{array}$ is a continuous function, then the value of $\stackrel{Lim}{x\to 1}\frac{tan(1-x)f(x-1)}{x^2+3x+2}$is.

• A. $\frac{-9}{2}$
• B. $\frac{9}{2}$
• C. $\frac{9}{4}$
• D. $\frac{-9}{4}$

Answer 1

The correct option is B $\frac{9}{2}$

$f\left(0\right)=K=\underset{x\to 0}{lim}\frac{1-cos3x}{x^2}=\frac{9}{2}$
$\therefore \underset{x\to 0}{lim}\frac{tan(1-x)f(x-1)}{x^2-3x+2}=\underset{x\to 0}{lim}\left(\frac{tan(1-x)}{(x-1)}\right)\left(\frac{f(x-1)}{x-2}\right)$
$=-1\times \left(\frac{-9}{2}\right)=\frac{9}{2}$