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

I get $$\lim_{x\to\infty} {x+1} = \infty$$ But I am wrong. Why?

With L'Hospital, I get $$\lim_{x\to\infty}\frac{1}{2}\sqrt{x^2 + 2x + 1} \cdot (2x+2) = \frac{1}{2} \sqrt{1} 2 = 1$$

• You applied L'Hôpital's rule incorrectly. – Michael Rybkin Sep 30 '19 at 7:31
• I'd like to say how much the notation $\frac{1}{2}\sqrt{1}2$ bothers me! :-P – Theo Bendit Sep 30 '19 at 7:41
• I am afraid you confused $\infty$ and $0$ (not counting that you did not apply L'Hospital at all). – Yves Daoust Sep 30 '19 at 7:57
• The question is not in a form where L'Hopital's rule can be used. – Sam Sep 30 '19 at 8:34

If $$x\geqslant-1$$, then $$\sqrt{x^2+2x+1}=x+1$$ and therefore, yes,$$\lim_{x\to\infty}\sqrt{x^2+2x+1}=\lim_{x\to\infty}x+1=\infty.$$

And you cannot apply L'Hopital's rule, since you do not have an indeterminate form here.

The correct answer is $$\begin{gathered} \mathop {\lim }\limits_{x \to \infty } \sqrt {x^2 + 2x + 1} = \hfill \\ \hfill \\ = \mathop {\lim }\limits_{x \to \infty } \sqrt {\left( {x + 1} \right)^2 } = \mathop {\lim }\limits_{x \to \infty } \left| {x + 1} \right| = + \infty \hfill \\ \end{gathered}$$ De L'Hospital's Rule is for limits of the type $$\mathop {\lim }\limits_{x \to \infty } \frac{{f(x)}} {{g(x)}}$$ with $$\mathop {\lim }\limits_{x \to \infty } f(x) = \infty$$ and $$\mathop {\lim }\limits_{x \to \infty } g(x) = \infty$$ under suitable hypothesis. This is not your case

Without L'Hopital's rule you can rewrite:

$$\lim_{x \to \infty} \sqrt{x^2 + 2x + 1} = \lim_{x \to \infty} \sqrt{(x + 1)^2} = \lim_{x \to \infty} |x+1|$$

Since $$x$$ is a large value, you can omit absolute value: $$\lim_{x \to \infty} (x+1)=\infty$$.

L'Hopital applies to the indeterminate forms $$\infty/\infty$$ or $$0/0$$.

In this case we have simply

$$\sqrt{x^2 + 2x + 1}\ge x\to \infty$$