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Good day, I'm now prepping for my calculus finals and reviewing some examples that I couldn't complete during the semester.

I'm wondering how I solve this (without L'Hôpital):

$$\lim_{x\to-\infty} \frac{-x}{\sqrt{4+x^2}}$$


$$\lim_{x\to 0^+} e^{-2/x^3}.$$

Any help is well appreciated.

PS. Can I append to this one question other questions or should I start another question for each problem?

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Re: P.S., the latter option is preferable. – J. M. Dec 17 '11 at 16:12
Thank, wasn't sure. – Siemsen Dec 17 '11 at 19:31
up vote 4 down vote accepted

$1$. We want to find $$\lim_{x\to-\infty}\frac{-x}{\sqrt{x^2+4}}$$ The tricky thing here is to avoid a sign error. Let's first find out what the answer should be. Imagine that $x$ is very large negative, like $-1000000$. Then the top is $1000000$. The bottom is very close to $1000000$. So the ratio is about $1$. One can get additional informal confirmation by "plugging in" various large negative values, using a calculator.

One way to carry out a formal manipulation with low risk of error is to take the cowardly way out and avoid negative numbers. Let $y=-x$. Then we want $$\lim_{y\to\infty} \frac{y}{\sqrt{y^2+4}}.$$ Divide top and bottom by $y$, and note that $\frac{\sqrt{y^2+4}}{y}=\sqrt{1+\frac{4}{y^2}}$. So we want $$\lim_{y\to\infty} \frac{1}{\sqrt{1+\frac{4}{y^2}}}.$$ Let $y$ get very large positive. The bottom approaches $1$, so the limit is $1$.

We could also work directly with $x$. Divide top and bottom by $x$. The top gives no problem, so now look at $\frac{\sqrt{x^2+4}}{x}$. We want to take the $x$ "inside."

It would be wrong to say that $\frac{\sqrt{x^2+4}}{x}=\sqrt{1+\frac{4}{x^2}}$. For note that we are interested in the expression when $x$ is negative. When $x$ is negative, $\frac{\sqrt{x^2+4}}{x}$ is negative but $\sqrt{1+\frac{4}{x^2}}$ is positive.

The correct assertion is that $$\frac{\sqrt{x^2+4}}{x}=-\sqrt{1+\frac{4}{x^2}}\qquad\text{if $x<0$}.$$ Now everything goes through just fine. When $x<0$, $$\frac{-x}{\sqrt{x^2+4}}=\frac{-1}{-\sqrt{1+\frac{4}{x^2}}}=\frac{1}{\sqrt{1+\frac{4}{x^2}}},$$ and finding the limit as $x\to-\infty$ is straightforward.

$2.$ We want $$\lim_{x\to 0^+}e^{-2/x^3}.$$ This one is straightforward. Imagine $x$ close to $0$ but positive. Then $-2/x^3$ is large negative, and therefore $e^{-2/x^3}$ is ridiculously close to $0$. Our limit is $0$.

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Thanks for the answer, much appreciated. Hopefully this will be my only question. – Siemsen Dec 17 '11 at 19:02

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