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

  1. $\frac{3}{5}$
  2. $\frac{5}{3}$
  3. $1$
  4. $\infty$
in Calculus by Boss (17.5k points)
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1 Answer

+1 vote
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.
by Boss (19.1k points)
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