Find $\lim_{x\to -\infty}\sqrt{x^2+9}+x+3$

$\lim_{x\to -\infty}\sqrt{x^2+9}+x+3$

The answer is 3. I know $\sqrt{x^2+9}$ gets infinitely big ($>0$) and so does $x$ negatively, so it seems reasonable to think that the sum of them will aproach $0$ as $x$ goes to infinity, but is that true for the limit of any function $f(x) + g(x)$ when $f(x)$ goes to $\infty$ and $g(x)$ goes to $-\infty$? I tried to prove that "lemma" myself but couldn't find a way through. If that's not always true it must certainly be true for at least the situation of the presented problem, but I couldn't manage to prove that in a consistent way. Any suggestions?

Solution 1:

write $$\sqrt{x^2+9}+x+3=\frac{-6x}{\sqrt{x^2+9}-x-3}$$

Solution 2:

HINT:

WLOG let $-1/x=h\implies h\to0^+$

$\sqrt{x^2+9}=\dfrac{\sqrt{1+9h^2}}{|h|}=\dfrac{\sqrt{1+9h^2}}h$ as $h>0$

$\lim_{x\to -\infty}\sqrt{x^2+9}+x+3=\lim_{h\to0^+}\dfrac{\sqrt{1+9h^2}-(1-3h)}h$

Now rationalize the numerator

Solution 3:

Let $x=-3\cot2h\implies h\to0^+$

$\lim_{x\to -\infty}\sqrt{x^2+9}+x+3=3+3\cdot\lim_{h\to0^+}(\csc2h-\cot2h)$

Now $\lim_{h\to0^+}(\csc2h-\cot2h)=\lim_{h\to0^+}\dfrac{1-\cos2h}{\sin2h}=\lim_{h\to0^+}\dfrac{2\sin^2h}{2\sin h\cos h}=?$