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

Question

I'm looking to evaluate

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

The answer in the book is $2$. How do I simply evaluate this problem?

I usually solve limits such as this with the short cut method, i.e (Numerator degree < Denominator degree) = 0 ; (Numerator degree = Denominator degree )= take ratio of leading coefficients; (Degree numerator > degree denominator )= take leading terms and use algebra to simplify and then plug in $-\infty$ or $+\infty$ Please keep in mind that I do not know L'Hopital's rule.

Have a look at this question and other questions linked to it. — Martin Sleziak, May 06 '15 at 17:12
It is really silly to call this post a duplicate. It looks nothing like the other post. Absurd. — Cetshwayo, May 08 '15 at 17:22

score 8 · Accepted Answer · answered May 06 '15 at 13:13

$$\begin{align}\lim_{x\to \infty}\sqrt{x^2+4x+1}-x&=\lim_{x\to\infty}(\sqrt{x^2+4x+1}-x)\cdot\frac{\sqrt{x^2+4x+1}+x}{\sqrt{x^2+4x+1}+x}\\&=\lim_{x\to\infty}\frac{(x^2+4x+1)-x^2}{\sqrt{x^2+4x+1}+x}\\&=\lim_{x\to\infty}\frac{4+\frac 1x}{\sqrt{1+\frac 4x+\frac{1}{x^2}}+1}\\&=\frac{4}{1+1}\end{align}$$

score 0 · Answer 2 · answered May 06 '15 at 13:13

0

Hint: $\sqrt{x^2+4x+1} = \sqrt{x^2+4x+4-3} = \sqrt{(x+2)^2-3} = \dots$

answered May 06 '15 at 13:13

Keba

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

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