# Prove the following

Question:

$\lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}$

Solution:

Given $\lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}$

Now to rationalize the denominator by multiplying the given equation by its rationalizing factor we get

$\Rightarrow \lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}=\lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1} \times\left(\frac{\sqrt{x}+1}{\sqrt{x}+1}\right)$

On simplifying the above equation we get

$\Rightarrow \lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1} \times\left(\frac{\sqrt{x}+1}{\sqrt{x}+1}\right)=\lim _{x \rightarrow 1} \frac{x^{4} \sqrt{x}+x^{4}-x-\sqrt{x}}{x-1}$

Taking $\sqrt{x}$ common we get

$\lim _{x \rightarrow 1} \frac{x^{4} \sqrt{x}-x^{4}-x-\sqrt{x}}{x-1}=\lim _{x \rightarrow 1} \frac{\sqrt{x}\left(x^{4}-1\right)+x\left(x^{3}-1\right)}{x-1}$

Using $a^{3}-b^{3}$ and $a^{2}-b^{2}$ formula and expanding we get

$\Rightarrow \lim _{x \rightarrow 1} \frac{\sqrt{x}\left(x^{4}-1\right)+x\left(x^{2}-1\right)}{x-1}=\lim _{x \rightarrow 1} \frac{\sqrt{x}(x-1)(x+1)\left(x^{2}+1\right)+x(x-1)\left(x^{2}+x+1\right)}{x-1}$

On simplification we get

$\Rightarrow \lim _{x \rightarrow 1} \frac{\sqrt{x}(x-1)(x+1)\left(x^{2}+1\right)+x(x-1)\left(x^{2}+x+1\right)}{x-1}=\lim _{x \rightarrow 1} \frac{(x-1)\left(\sqrt{x}(x+1)\left(x^{2}+1\right)+x\left(x^{2}+x+1\right)\right)}{x-1}$

$\Rightarrow \lim _{x \rightarrow 1}\left(\sqrt{x}(x+1)\left(x^{2}+1\right)+x\left(x^{2}+x+1\right)\right)=4+3=7$

$\lim _{x \rightarrow 1}\left(\sqrt{x}(x+1)\left(x^{2}+1\right)+x\left(x^{2}+x+1\right)\right)=4+3=7$

$\Rightarrow \lim _{x \rightarrow 1} \frac{x^{4}-\sqrt{x}}{\sqrt{x}-1}=7$