ISI2015-DCG-52

Question

$\underset{x \to -1}{\lim} \dfrac{1+\sqrt[3]{x}}{1+\sqrt[5]{x}}$ equals

• A. $\frac{3}{5}$
• B. $\frac{5}{3}$
• C. $1$
• D. $\infty$

Answer 1

Answer: $\mathbf B$

Solution:

Let $$\mathrm{ I = \underset{x\rightarrow-1}\lim\frac{1+\sqrt[3]{x}}{1+\sqrt[5]{x}}}\ \tag{1}$$

Applying L'Hospital's Rule in $(1)$, we get:

 $$\mathrm{\Rightarrow I=\underset{x\rightarrow-1}\lim\frac{\frac{d}{dx}\left(1+\sqrt[3]{x}\right)}{\frac{d}{dx}\left(1+\sqrt[5]{x}\right)}}$$

$$\mathrm{=\underset{x\rightarrow-1}\lim\frac{\frac{d}{dx}\left(1+x^{\frac{1}{3}}\right)}{\frac{d}{dx}\left(1+x^{\frac{1}{5}}\right)}}$$

$$\mathrm{=\underset{x\rightarrow-1}\lim\frac{\frac{1}{3}x^{\left(1-\frac{1}{3}\right)}}{\frac{1}{5}x^{\left(1-\frac{1}{5}\right)}}} $$

$$\mathrm{= \underset{x\rightarrow-1}\lim\frac{\frac{1}{3}x^{\frac{2}{3}}}{\frac{1}{5}x^{\frac{4}{5}}}}$$

$$\mathrm{=\underset{x\rightarrow-1}\lim\frac{\frac{1}{3}x^{2.{\frac{1}{3}}}}{\frac{1}{5}x^{4{.\frac{1}{5}}}}}$$

Now, substitute $\mathbf {x = -1}$ above, $\because \mathbf {x^2\;\text{and} \; x^4}$ terms have even powers and anything raised to power $\mathbf 1$ is $\mathbf 1$, so the equation reduces to:

$$\mathrm I= \dfrac{\frac{1}{3}} {\frac{1}{5}} = \dfrac{5}{3}$$

$\therefore \mathbf B$ is the correct option.