The value of lim _{x rightarrow 1} frac{x^4-s | vedclass

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(B) Let $L = \lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$.Since the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital's$ rule:$

Vedclass · Accepted Answer

$7$

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(B) Let $L = \lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$.Since the limit is in the $\frac{0}{0}$ form,we apply $L'H\hat{o}pital's$ rule:$L = \lim _{x \rightarrow 1} \frac{\frac{d}{dx}(x^4-\sqrt{x})}{\frac{d}{dx}(\sqrt{x}-1)}$$L = \lim _{x \rightarrow 1} \frac{4x^3 - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$L = \lim _{x}$ ${\rightarrow 1} \left( \frac{4x^3}{\frac{1}{2\sqrt{x}}} - \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}} \right)$$L = \lim _{x \rightarrow 1} (8x^3\sqrt{x} - 1)$$L = 8(1)^3\sqrt{1} - 1 = 8 - 1 = 7$.

The value of $\lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$ is

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