Answer: $B$
Given: $$\underset{x\rightarrow-1}\lim\frac{1+\sqrt[3]{x}}{1+\sqrt[5]{x}}\qquad \to(1)$$
Applying L'Hospital's Rule in $(1)$, we get:
$$=\underset{x\rightarrow-1}\lim\frac{\frac{dy}{dx}\Big(1+\sqrt[3]{x}\Big)}{\frac{dy}{dx}\Big(1+\sqrt[5]{x}\Big)}$$
$$=\underset{x\rightarrow-1}\lim\frac{\frac{dy}{dx}\Big(1+x^{\frac{1}{3}}\Big)}{\frac{dy}{dx}\Big(1+x^{\frac{1}{5}}\Big)}$$
$$=\underset{x\rightarrow-1}\lim\frac{\frac{1}{3}x^{\Big(1-\frac{1}{3}\Big)}}{\frac{1}{5}x^{\Big(1-\frac{1}{5}\Big)}} $$
$$= \underset{x\rightarrow-1}\lim\frac{\frac{1}{3}x^{\frac{2}{3}}}{\frac{1}{5}x^{\frac{4}{5}}}$$
$$=\underset{x\rightarrow-1}\lim\frac{\frac{1}{3}x^{2.{\frac{1}{3}}}}{\frac{1}{5}x^{4{.\frac{1}{5}}}} $$
Now, substitute $x = -1$ above, since $x^2\;and \; x^4$ terms have even powers and anything raised to power $1$ is $1$, so the equation reduces to:
$$= \frac{\frac{1}{3}} {\frac{1}{5}} = \frac{5}{3}$$
$\therefore B$ is the right answer.