Find $\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$ $$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$$
I have tried multiplying by $\frac{1}{\sqrt{x^2+4}}$ and it's reciprocal, but I cannot seem to find the solution. L'Hospital's doesn't seem to work either, as I keep getting rational square roots.
 A: Hint $$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt {{x^2} + 4} }}{{x + 4}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\dfrac{{\sqrt {{x^2} + 4} }}{x}}}{{\dfrac{{x + 4}}{x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt {\dfrac{{{x^2} + 4}}{{{x^2}}}} }}{{1 + \dfrac{4}{x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + \dfrac{4}{{{x^2}}}} }}{{1 + \dfrac{4}{x}}}$$
A: $${\sqrt{x^2+4}\over x+4}={x\over x+4}\sqrt{1+(4/x^2)}$$ Can you take it from there?
A: Another way to look at it is to make use of the following inequality. $$x \leq \sqrt{x^2+4} \leq x+2, \,\, \forall x \in \mathbb{R}^+$$
Hence, we have that $$\dfrac{x}{x+4} \leq \dfrac{\sqrt{x^2+4}}{x+4} \leq \dfrac{x+2}{x+4}$$
Now apply the squeeze/sandwich theorem to get $$\lim_{x \to \infty}\dfrac{\sqrt{x^2+4}}{x+4} = 1$$
A: $\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$=$\lim\limits_{x \to \infty} \frac{\sqrt{x^2(1 + \frac{4}{x^2})}}{x+4}$=$\lim\limits_{x \to \infty} \frac{x\sqrt{1 + \frac{4}{x^2}}}{x+4}$.
When $ x\longrightarrow\propto $   $\Rightarrow$     $\frac{4}{x^2}$$\longrightarrow$ 0 .
The above integral then takes the form:
$\lim\limits_{x \to \infty} \frac{x}{x+4}$=$\lim\limits_{x \to \infty} \frac{x}{x(1+\frac{4}{x})}$.=$\lim\limits_{x \to \infty} \frac{1}{1+\frac{4}{x}}$.
When $ x\longrightarrow\propto $  $\Rightarrow$  $\frac{4}{x}$$\longrightarrow$ 0 .
Now the above integral then takes the form:
$\lim\limits_{x \to \infty} \frac{1}{1+\frac{4}{x}}$=$\lim\limits_{x \to \infty} \frac{1}{1+0}$=1.
$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$=1
A: HINT:
$$\left(\frac{\sqrt{x^2\:+\:4}}{x+4}\right)=\left(\frac{\left|x\right|\sqrt{1\:+\:\frac{4}{x^2}}}{x+4}\right)$$
So
$$\lim _{x\to +\infty }\left(\frac{\left|x\right|\sqrt{1\:+\:\frac{4}{x^2}}}{x+4}\right) = 1$$
$$\lim _{x\to -\infty }\left(\frac{\left|x\right|\sqrt{1\:+\:\frac{4}{x^2}}}{x+4}\right) = -1$$
A: Hint:
In the numerator, the dominant term of the polynomial is $x^2$, of which you take the square root. In the denominator, the dominant term is $x$. Then the function behaves like $|x|/x$ at infinity.

Or
$$\lim_{x\to\infty}\frac{\sqrt{x^2+4}}{x+4}=\lim_{x\to\infty}\frac{\sqrt{(x-4)^2+4}}x=\lim_{x\to\infty}\sqrt{1-\frac{8}{x}+\frac{20}{x^2}}.$$
