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

1. $\frac{3}{5}$
2. $\frac{5}{3}$
3. $1$
4. $\infty$
in Calculus
recategorized | 32 views

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.

by Boss (19.2k points)
edited by

+1 vote