# How to calculate the limit $\lim_{x\to + \infty} \frac{\sqrt{x^2+4}}{x}$?

So I have another question on limits. In this case the limit is:

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

I tried to transform it into $$\lim_{x\to + \infty} \frac{x^2+4}{x^2}$$ because I wanted to solve it by factoring but I checked the graph of both and they are different..

I'm stuck here cause my idea doesn't work and I don't know how to solve it! Could someone be so kind to help me? Thank you in advance! :)

• Though note that you're still getting the same answer if you take the positive square root of the limit you're getting in your squared expression. Commented Dec 31, 2017 at 15:45
• Informally, as $x$ gets larger, the $4$ contributes less, percent-wise, to the value of $x^2+4$. So you can think of $\sqrt{x^2+4}$ as just a bit bigger than $x$, say $x + \delta$ where $\frac{\delta}{x}$ gets smaller as $x$ gets bigger. Then $\frac{\sqrt{x^2+4}}{x} \to \frac{x + \delta}{x} \to 1 + \frac{\delta}{x} \to 1$. Commented Dec 31, 2017 at 16:27

Factoring $x^2$ out of the expression under the root, you get : $$\lim_{x\to + \infty} \frac{\sqrt{x^2+4}}{x} = \lim_{x\to + \infty} \frac{\sqrt{x^2(1+4/x^2)}}{x} = \lim_{x\to + \infty} \frac{|x|\sqrt{1+4/x^2}}{x}$$ Now, note that $x\to +\infty$ which means that $x>0$, thus :

$$\lim_{x\to + \infty} \frac{|x|\sqrt{1+4/x^2}}{x}= \lim_{x\to + \infty} \frac{x\sqrt{1+4/x^2}}{x}=\lim_{x\to + \infty}\sqrt{1+4/x^2}=1$$

\begin{align}\lim_{x\to+\infty}\frac{\sqrt{x^2+4}}x&=\lim_{x\to+\infty}\sqrt{\frac{x^2+4}{x^2}}\\&=\sqrt{\lim_{x\to+\infty}1+\frac4{x^2}}\\&=\sqrt{1}\\&=1.\end{align}

Let $x>0$.

$1= \dfrac{\sqrt{x^2}}{x} \lt \dfrac{\sqrt{x^2+4}}{x} \lt$

$\dfrac{(x+2)}{x}= 1+ \dfrac{2}{x}.$

And the limit is?

Used: $x^2+4 \lt (x+2)^2$, for $x >0$.

• Peter you don't bound to 1 in such way.
– user
Commented Dec 31, 2017 at 15:49
• You're missing something on the RHS in the first inequality. I'm not sure if it was intentionally left blank as a hint :/ Commented Dec 31, 2017 at 15:52
• $\dfrac{\sqrt{x^2}}{x} \lt \dfrac{\sqrt{x^2+4}}{x} \lt \dfrac{(x+2)}{x}= 1+ \dfrac{2}{x}$
– user
Commented Dec 31, 2017 at 15:54
• Thanks, gimusi, I got it, as frequently, I miss half the answer, my specialty:))). Commented Dec 31, 2017 at 16:10
• You idea is very good Peter, complete the answer!
– user
Commented Dec 31, 2017 at 16:12

It occurred to me that the problem would be a lot easier if $\sqrt{x^2+4}$ was $\sqrt{x^2+4}-x$, So I came up with this.

\begin{align} \frac{\sqrt{x^2+4}}{x} &= \left(\frac{\sqrt{x^2+4}}{x}-1\right) + 1 \\ &=\frac{\sqrt{x^2+4}-x}{x} + 1 \\ &=\frac{(\sqrt{x^2+4}-x)(\sqrt{x^2+4}+x)}{x(\sqrt{x^2+4}+x)} + 1 \\ &=\frac{4}{x(\sqrt{x^2+4}+x)} + 1 \\ &\to 1 \ \text{as} \ x \to \infty \end{align}

Since $f(x)=\sqrt{x}$ is a continuous functuion, we obtain: $$\lim_{x\rightarrow+\infty}\frac{\sqrt{x^2+4}}{x}=\lim_{x\rightarrow+\infty}\frac{x\sqrt{1+\frac{4}{x^2}}}{x}=\sqrt{\lim\limits_{x\rightarrow+\infty}\left(1+\frac{4}{x^2}\right)}=1$$