Evaluate the following limit: Show that _ x ( x^2+x+1 -x ) _ x ( x^2+1 -x )

R.H.S= _ x ( x^2+1 -x ) = _ x ( x^2+1 -x ) ( x^2+1 +x ) ( x^2+1 -x ) = _ x x^2+1 -x x 1+1 x^2 +x = _ x 1 x 1+1 x^2 +x = _ x 1 x 1 x 1+1 x^2 +x =0 Also, L.H.S= _ x ( x^2+x+1 -x ) = _ x ( x^2+x+1 -x ) ( x^2+x+1 +x ) ( x^2+x+1 +x ) = _ x ( x^2+x+1 -x^2 ) x^2+x+1 +x = _ x x (1+1 x ) x (1+1 x +1 x^2 +1 ) = _ x 1+1 x 1+1 x +1 x^2 +1 =1 1+1 = 12 _ x ( x^2+x+1 -x ) is not equal to _ x ( x^2+1 -x ).

Evaluate the following limit: Show that _ x ( x^2+x+1 -x ) _ x ( …

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Question

Evaluate the following limit:
Show that $\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)\ne\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$

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Answer

$\text{R.H.S=}\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+1}-\text{x}\big)$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\sqrt{\text{x}^2+1}-\text{x}\big)\big(\sqrt{\text{x}^2+1}+\text{x}\big)}{\big(\sqrt{\text{x}^2+1}-\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{{\text{x}^2+1}-\text{x}}{\text{x}\sqrt{1+\frac{1}{\text{x}^2}}+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1}{\text{x}\sqrt{1+\frac{1}{\text{x}^2}}+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1}{\text{x}}\begin{pmatrix}\frac{1}{\text{x}\sqrt{1+\frac{1}{\text{x}^2}}+\text{x}}\end{pmatrix}$
$=0$
Also,
$\text{L.H.S}=\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)\big(\sqrt{\text{x}^2+\text{x}+1}+\text{x}\big)}{\big(\sqrt{\text{x}^2+\text{x}+1}+\text{x}\big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}^2\big)}{\sqrt{\text{x}^2+\text{x}+1}+\text{x}}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{\text{x}\Big(1+\frac{1}{\text{x}}\Big)}{\text{x}\Big(1+\frac{1}{\text{x}}+\frac{1}{\text{x}^2}+1\Big)}$
$=\lim\limits_{\text{x}\rightarrow\infty}\frac{1+\frac{1}{\text{x}}}{\sqrt{1+\frac{1}{\text{x}}+\frac{1}{\text{x}^2}}+1}$
$=\frac{1}{1+1}=\frac12$
$\therefore\ \lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+\text{x}+1}-\text{x}\big)$ is not equal to $\lim\limits_{\text{x}\rightarrow\infty}\big(\sqrt{\text{x}^2+1}-\text{x}\big).$

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