Evaluate the following limit: _ x 0 1+x^2 - 1+x 1+x^3 - 1+x

Question

Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}}{\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}}$

✓

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}}{\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}}$$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}-\sqrt{1+\text{x}}\big)}\times\frac{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(1+\text{x}^2\big)-\big(1+\text{x}\big)\times\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^3}-\sqrt{1+\text{x}}\big)\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\text{x}^2-\text{x}\big)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\big(1+\text{x}^3-1-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{x}-1)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)\times\big(\text{x}^2-1\big)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}(\text{x}-1)\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)(\text{x})(\text{x}-1)(\text{x}+1)}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\big(\sqrt{1+\text{x}^3}+\sqrt{1+\text{x}}\big)}{\big(\big(\sqrt{1+\text{x}^2}+\sqrt{1+\text{x}}\big)(\text{x}+1)\big)}$
$=\frac{2}{2}=1$