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I am studying limits at infinity, and I have a doubt about evaluating them. From what I know, limits only exist if both sides of the limit exist and are equal. For example, take a look at the following limit:

$$\lim_{ x\to \infty}\frac{x}{e^{-x}}$$

As $x \to +\infty$ the function goes to $+\infty$. However, as $x \to -\infty$ the function goes to $0$.

Does this mean that the limit does not exist or I only have to evaluate the limit as x goes to $+\infty$?

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    $\begingroup$ The limit at positive infinity is a different problem then the limit at negative infinity. They don't have to be the same and very often they aren't the same, like in your example. One limit exist (second one) and the other doesn't (first one). I think you are confused with taking the limit at a single point, where right hand limit has to be the same as left hand limit in order for THE limit to exist. $\endgroup$ – imranfat Aug 30 '17 at 0:47
  • $\begingroup$ That means that I have to evaluate it with x approaching plus infinity? because that problem does not specify if it is plus infinity or minus infinity. $\endgroup$ – Nick_17 Aug 30 '17 at 0:52
  • $\begingroup$ in that case i believe the problem is not well formulated $\endgroup$ – Marios Gretsas Aug 30 '17 at 0:52
  • $\begingroup$ When looking purely at the limit as you presented, you only consider the limit as $x$ goes to infinity. That answer is infinity and so technically that limit does not exist although some people take infinity as an acceptable answer, as Marios does. $\endgroup$ – imranfat Aug 30 '17 at 0:54
  • $\begingroup$ @inframat a limit which has an infinite value is an existent limit..in another case if $x$ approaches a point from the left and right and you have different limits then the limit there does not exist..take for instance $\frac{1}{x}$ as $x \to 0$ $\endgroup$ – Marios Gretsas Aug 30 '17 at 1:01
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Although closely related:

i) $\lim_{x\rightarrow a}f(x) = L; a\in \mathbb R$

ii) $\lim_{x\to a^+}f(x) = L$

iii) $\lim_{x\to a^-}f(x) = L$

iv) $\lim_{x\to \infty} f(x) = L$

and v) $\lim_{x\to -\infty} f(x) = L$.

have technically different definitions.

i) means as $x$ gets close to $a$, then $f(x)$ gets close to $L$. or technically:

i) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $|x -a| < \delta$ it will follow that $|f(x) - L| < \epsilon$.

ii) means as $x$ gets close to $a$ but larger than $a$, then $f(x)$ gets close to $L$. or technically:

ii) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $a < x < a+ \delta$ it will follow that $|f(x) - L| < \epsilon$.

Notice that if i) is true than ii) must be true but if ii) is true i) is not nescessarily true.

iii) means as $x$ gets close to $a$ but smaller than $a$, then $f(x)$ gets close to $L$. or technically:

iii) means For any $\epsilon > 0$ we can find $\delta$ so that whenever $a -\delta < x < a$ it will follow that $|f(x) - L| < \epsilon$.

Notice that if i) is true than ii) and iii) must be true. And if both ii) and iii) is true then i) is true. But if one or the other of ii) or iii) is true but the other isn't i) will not be true.

So the statement "i) if ii) and iii)" is not a definition but an observation. Well.... it could be a definition as they are equivalent.

Now...

Limit to infinity is not that as $x$ gets close to infinity $f(x)$ will get close to $L$. That wouldn't make any sense because we can't get $x$ "close to infinity". Instead we mean, as $x$ becomes large, $f(x)$ gets close to $L$. or technically:

iv) For any $\epsilon > 0$ there is some $M \in \mathbb R$ so that whenever $x > M$ it will follow that $|f(x) - L| < \epsilon$.

It makes no sense to talk of $\lim_{x\to \infty^+} f(x)$ because we can't "approach from the right". And to talk of $\lim_{x \to \infty^-}f(x)$ is the exact same thing as $\lim_{x \to \infty}f(x)$ because we can only "approach from the left".

So it is NOT the case that $\lim_{x\to\infty} f(x) = L$ if $\lim_{x\to \infty^+}f(x) = L$ and $\lim_{x\to \infty^-}f(x) = L$. Instead it is just $\lim_{x\to\infty} f(x) = L$ if $\lim_{x\to \infty^-}f(x) = L$.

$\lim_{x \to -\infty}f(x)=L$ means we are talking the values of very big negative numbers (very small, or negative with very large magnitude or absolute values). It should be very clear that $-\infty$ means infinity beyond the negative extreme of the reals, whereas $\infty^-$ means something entirely different; it means approaching the the positive extreme of the real numbers but at some finite value less than infinity.

v) For any $\epsilon > 0$ there is some $M \in \mathbb R$ so that whenever $x < M$ it will follow that $|f(x) - L| < \epsilon$.

Again it makes no sense to talk of $\lim_{x\to -\infty^-}f(x)$ and it is redundant to talk of $\lim_{x\to -\infty^+} f(x)$ for the exact same reasons.

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$$\frac{x}{e^{-x}}=xe^x$$

If you take $x \to +\infty$ then the limit is $+ \infty$

If you take $x \to -\infty$ then the limit is $0$

You can say that the limit exist because the existence of the limit as $x$ goes to $-\infty$ does not depend to the limit as $x$ goes to $+\infty$

In both cases the limit exists,just in the first case is $+\infty$

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  • $\begingroup$ So, do I have to evaluate both cases and thus give two answers? $\endgroup$ – Nick_17 Aug 30 '17 at 0:58
  • $\begingroup$ as i said in the above comment in this case the problem is not well formulated..the author must be more specific $\endgroup$ – Marios Gretsas Aug 30 '17 at 1:00
  • $\begingroup$ Also @Nick_17..some authors consider an infinite limit existent and some others not ...i don't think personally that one these two cases is wrong $\endgroup$ – Marios Gretsas Aug 30 '17 at 1:07
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This graph might help you see what is going on.

enter image description here

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