<s>[INST] 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.
 [/INST] $\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$=$\lim\limits_{x \to \infty} \frac{\sqrt{x^2(1 + \frac{4}{x^2})}}{x+4}$=$\lim\limits_{x \to \infty} \frac{x\sqrt{1 + \frac{4}{x^2}}}{x+4}$.
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
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