Question

If $g\left(x\right)=\{\begin{array}{cc}\frac{a{e}^{1/|x+2|}-1}{2-e^{1/|x+2|}}& -3<x<-2\\ b& x=-2\\ \sin \left(\frac{x^4-16}{x^5+32}\right)& -2<x<0\end{array}$ is continuous at $x=-2$, then

• A. $a=\sin \left(\frac{3}{5}\right)$
• B. $a=-\sin \left(\frac{2}{5}\right)$
• C. $a=\sin \left(\frac{2}{5}\right)$
• D. $b=-a$

Answer 1

Answer: (B), (D)

$b=-a$
$a=-\sin \left(\frac{2}{5}\right)$

$\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}\frac{a{e}^{1/|x+2|}-1}{2-e^{1/|x+2|}}$

$=\underset{x\to 2^{-}}{lim}\frac{a{e}^{-1/|x+2|}-1}{2e^{-1/|x+2|}-1}=-a$

$(\frac{1}{|x+2|}\to \infty asx\to -2^{-})$

$\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to -2^{+}}{lim}\sin \left(\frac{x^4-16}{x^5+32}\right)$

$=\underset{x\to 2^{+}}{lim}\sin \frac{(x-2)(x^2+4)}{(x^4-2x^3+4x^2-8x+16)}$

$=\sin \frac{(-4)\left(8\right)}{16+16+16+16+16}$

$=\sin (-\frac{2}{5})$