_ x 0 ( 1+x^2-1+x 1+x^3-1+x )=? - Vidyadip

Question

$\lim _{x \rightarrow 0}\left(\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right)=?$

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Answer

$\lim _{x \rightarrow 0}\left(\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right)$
$=\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}} \times \frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}\right.$$\left.\times \frac{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}{\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\right\}$
$\begin{array}{l}=\lim _{x \rightarrow 0}\left\{\frac{\left(1+x^2\right)-(1+x)}{\left(1+x^3\right)-(1+x)} \times \frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right\} \\ =\lim _{x \rightarrow 0}\left\{\frac{x(x-1)}{\left(x^3-x\right)} \times \frac{\sqrt{1+x^3}+\sqrt{1+x}}{\sqrt{1+x^2}+\sqrt{1+x}}\right\}\end{array}$
$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{x(x-1)\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{x(x-1)(x+1)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)} \quad(x \neq 0) \\ =\lim _{x \rightarrow 0} \frac{\left(\sqrt{1+x^3}+\sqrt{1+x}\right)}{(x+1)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}\end{array}$
$\begin{array}{l}=\frac{\left(\sqrt{1+0^3}+\sqrt{1+0}\right)}{(0+1)\left(\sqrt{1+0^2}+\sqrt{1+0}\right)} \\ =\frac{2}{1 \times 2} \\ =1\end{array}$

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