Question

$\lim_{x\to \infty} \frac{\sqrt{9x^{2} +x+1} -\sqrt{4x^{2} +2x+1} }{x+1}$

Answer 1

Answer:

$\lim _{x\to \infty \:}\left(\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\right)=1$

Step-by-step explanation:

Considering the expression

$\lim _{x\to \infty \:}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}$

Steps to solve

$\lim _{x\to \infty \:}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}$

$\mathrm{Divide\:by\:highest\:denominator\:power:}\:\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}$

$\lim _{x\to \infty \:}\left(\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}\right)$

$\lim _{x\to a}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)},\:\quad \lim _{x\to a}g\left(x\right)\ne 0$

$\mathrm{With\:the\:exception\:of\:indeterminate\:form}$

$\frac{\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)}{\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)}.....[1]$

As

$\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)=1$

Solving

$\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)....[A]$

$\lim _{x\to a}\left[f\left(x\right)\pm g\left(x\right)\right]=\lim _{x\to a}f\left(x\right)\pm \lim _{x\to a}g\left(x\right)$

$\mathrm{With\:the\:exception\:of\:indeterminate\:form}$

$\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}\right)-\lim _{x\to \infty \:}\left(\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)$

Also

$\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}\right)=3$

Solving

$\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}\right)......[B]$

$\lim _{x\to a}\left[f\left(x\right)\right]^b=\left[\lim _{x\to a}f\left(x\right)\right]^b$

$\mathrm{With\:the\:exception\:of\:indeterminate\:form}$

$\sqrt{\lim _{x\to \infty \:}\left(9+\lim _{x\to \infty \:}\left(\frac{1}{x}+\lim _{x\to \infty \:}\left(\frac{1}{x^2}\right)\right)\right)}$

$\lim _{x\to \infty \:}\left(9\right)=9$

$\lim _{x\to \infty \:}\left(\frac{1}{x}\right)=0$

$\lim _{x\to \infty \:}\left(\frac{1}{x^2}\right)=0$

So, Equation [B] becomes

⇒ $\sqrt{9+0+0}$

⇒ $3$

Similarly, we can find

$\lim _{x\to \infty \:}\left(\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)=2$

So, Equation [A] becomes

⇒ $3-2$

⇒ 1

Also

$\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)=1$

Thus, equation becomes

$\frac{\lim _{x\to \infty \:}\left(\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}\right)}{\lim _{x\to \infty \:}\left(1+\frac{1}{x}\right)}=\frac{1}{1}=1$

Therefore,

$\lim _{x\to \infty \:}\left(\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\right)=1$

Keywords: limit

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