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

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?

hululu
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R. Maia
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4 Answers4

1

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

Dr. Sonnhard Graubner
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1

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

lab bhattacharjee
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  • Thank you. How did you come up with that idea? Could you describe your line of thought? – R. Maia Apr 02 '17 at 14:35
  • @R.Maia, Setting the limit $\to0$ has reduced many complexities for me. See also: http://math.stackexchange.com/questions/2079417/why-is-lim-limits-x-to-infty-frac1-sqrtx22x-x-equal-to-0/2079432#2079432 and http://math.stackexchange.com/questions/2056840/solving-lim-x-to-infty-sqrtx2x1-sqrt1-xx2-using-o-littl/2056928#2056928 – lab bhattacharjee Apr 02 '17 at 14:38
1

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

lab bhattacharjee
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Since $x \to -\infty$ , $x^2 \to \infty$ and therefore $$x^2+9 \approx x^2$$

Therefore : $$\lim_{x \to -\infty}\sqrt{x^2+9}+x+3 \approx \lim_{x \to -\infty}\sqrt{x^2}+x+3 = \lim_{x \to -\infty}|x| +x+3$$

Since $x$ in negative, $|x| = -x$ ,Hence; $$|x| +x+3 = -x+x+3= \boxed{3}$$

Jaideep Khare
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