Question

If $\underset{x\to {-2}^{-}}{lim}\frac{a{e}^{1/|x+2|}-1}{2-e^{1/|x+2|}}=\underset{x\to {-2}^{+}}{lim}\sin \left(\frac{x^4-16}{x^5+32}\right)$, then $a$ is

• A. $-\sin \frac{3}{5}$
• B. $2$
• C. $-\sin \frac{2}{5}$
• D. $\sin \frac{1}{5}$

Answer 1

The correct option is C $-\sin \frac{2}{5}$

Given,
$\underset{x\to {-2}^{-}}{lim}\frac{a{e}^{1/|x+2|}-1}{2-e^{1/|x+2|}}=\underset{x\to {-2}^{+}}{lim}\sin \left(\frac{x^4-16}{x^5+32}\right)$

Now, $\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|}}{2e^{-1/|x+2|}-1}=-a$
Now, $=\underset{x\to -2^{+}}{lim}\sin \left(\frac{x^4-16}{x^5+32}\right)$
$=\underset{x\to -2^{+}}{lim}\sin \left(\frac{\frac{x^4-(-2{)}^4}{x-(-2)}}{\frac{x^5-(-2{)}^5}{x-(-2)}}\right)=\sin \left(\frac{4(-2{)}^3}{5(-2{)}^4}\right)=-\sin (\frac{2}{5})$
$\Rightarrow a=-\sin \frac{2}{5}$