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When I see a simple limit question like "find the limit as $x \rightarrow \infty$ when $f(x) =$ $3x+7 \over x+2$ I know that all I have to do is factor out x which makes $3 \over 1$ $\cdot$ $7 \over x$ / $2 \over x$, which both turn into 0 as I plug ONLY positive $\infty$ in for x.

But for limits like $e^x \over e^x +1$, as $x \rightarrow \infty$, how do I know that I have to plug in negative $\infty$ and positive $\infty$ for x after I simplify? (which gives me $1 \over 1 + (1/ e^x)$)

Does it have something to do with squeeze theorem? Thanks!

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    $\begingroup$ It the limit is $\lim x \to \infty$ that is a limit for positive large numbers as $x$ increases so $x$ is always positive. You never have to consider negative large numbers. Unless the limit is $\lim x \to -\infty$ in which cause you always check negative and never check positive. $\endgroup$ – fleablood Oct 27 '18 at 3:51
  • $\begingroup$ You aren't supposed to turn $+$ into $\cdot$ as you factor out $x$. $\endgroup$ – kasperd Oct 27 '18 at 9:34
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So a few things to sort out. First of all, no nothing to do with the squeeze theorem. What we can do with the squeeze theorem is say that when we can squeeze a limit between two other limits (or a constant and a limit) whose values we know coincide, then the limit we "squeezed" must agree with that as well.

As far as plugging in $+ \infty$ and $-\infty$ we never do that. We only take a limit going to a place. $+ \infty$ or $-\infty$ for example but not both.

What I think you're asking is "how do I know when I have to verify the limit from both the right and the left?" And the answer is, essentially, always. It's just that much of the time our function behaves nicely enough that it's very clear the left and right limits agree. When dealing with limits at $\pm \infty$ then there is only ever a one sided limit. How do you approach $+ \infty$ from the right for example? Well, you can't...

So for your example I'll abuse notation a bit and we have

$$\lim_{X \to \infty} \frac{e^x}{e^x+1}=\lim_{x \to \infty} \frac{1}{1+\frac{1}{e^x}}=\frac{1}{1+\frac{1}{e^\infty}}=\frac{1}{1+\frac{1}{\infty}}=\frac{1}{1+0}=1$$

$$\lim_{X \to -\infty} \frac{e^x}{e^x+1}=\lim_{x \to -\infty} \frac{1}{1+\frac{1}{e^x}}=\frac{1}{1+\frac{1}{e^{-\infty}}}=\frac{1}{1+e^\infty}=\frac{1}{1+\infty}=0$$

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  • $\begingroup$ and why does this equal 0 when $x \rightarrow - \infty$? $\endgroup$ – ming Oct 27 '18 at 4:35
  • $\begingroup$ he explained it rather well at the end, there. $\endgroup$ – The Count Oct 27 '18 at 4:40
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If you are told to find $\lim\limits_{x\to\infty}f(x)$, then you do not need to consider the case $x\to-\infty$.

You may be confused between these limits, and finding horizontal asymptotes of a function. To find a horizontal asymptote for a function $f(x)$, you need to check if $\lim\limits_{x\to\infty}f(x)$ exists. If it does, it equals some number $L$, and $y=L$ is a horizontal asymptote for $f(x)$. It is prudent in these situations to also to check $\lim\limits_{x\to-\infty}f(x)$, because some functions might have a horizontal asymptote distinct from the first one we found. $f(x)=\tan^{-1}x$ is an example.

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  • $\begingroup$ Sorry yes I was trying to find the horizontal asymptote but the way you do that is by finding the limit (I think?) so I just asked about the limit sorry. Anyways, for my example above for positive, I get why it equals 1, but how does making x approach negative $\infty$ make it 0? $\endgroup$ – ming Oct 27 '18 at 4:37
  • $\begingroup$ As $x\to\infty$, then $e^x\to\infty$, but as $x\to-\infty$, $e^x\to 0$. $\endgroup$ – The Count Oct 27 '18 at 4:38
  • $\begingroup$ but if $e^x \rightarrow 0$ then wouldn't $1 \over 1 + 1/0$ be the case, which is undefined? $\endgroup$ – ming Oct 27 '18 at 4:47
  • $\begingroup$ Use the original form of $\frac{e^x}{1+e^x}$. $\endgroup$ – The Count Oct 30 '18 at 3:17

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