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

Question

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

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

Answer 1

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.

Answer 2

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: $\lim_{x \to \infty} (x+1)=\infty$.

Answer 3

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

In this case we have simply

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

Answer 4

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