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! :)
 A: \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}
A: 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 $x >0$.
A: 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}
A: 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 : 
$$\lim_{x\to + \infty} \frac{|x|\sqrt{1+4/x^2}}{x}= \lim_{x\to + \infty} \frac{x\sqrt{1+4/x^2}}{x}=\lim_{x\to + \infty}\sqrt{1+4/x^2}=1$$
A: Since $f(x)=\sqrt{x}$ is a continuous functuion, we obtain:
$$\lim_{x\rightarrow+\infty}\frac{\sqrt{x^2+4}}{x}=\lim_{x\rightarrow+\infty}\frac{x\sqrt{1+\frac{4}{x^2}}}{x}=\sqrt{\lim\limits_{x\rightarrow+\infty}\left(1+\frac{4}{x^2}\right)}=1$$
