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If $\displaystyle t = \frac{1}{x}$ then

a) Explain why $\displaystyle\lim_{x \to 0^-}f(x)$ is equivalent to $\displaystyle\lim_{t \to -\infty}f\left(\frac{1}{t}\right)$

b) Using that, rewrite $\displaystyle\lim_{x \to 0^-} e^\frac{1}{x}$ as equivalent limit involving $t$ only. Evaluate the corresponding limit.

c) Use a similar technique to evaluate $\displaystyle\lim_{x \to \pm\infty} \ln\left(\frac{2}{x^3}\right)$

On a, I thought that they are equivalent because $f(x)$ is the inverse of $t$, is that correct?

On b, I get $\displaystyle\lim_{x \to 0^-} e^t$. How would I solve this limit? Isn't it the $t$'th root of $e$? Im confused...

And c just confuses me to begin with. How would I calculate that limit? I don't know where to start on that one and just need some help with getting my mind around it.


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2 Answers 2

up vote 4 down vote accepted

(a) does not have to do with $f$; rather, the point is that if $t=\frac{1}{x}$, then $x=\frac{1}{t}$, and $\lim\limits_{t\to-\infty}x = 0^-$ (that is, as $t$ approaches $-\infty$, the value of $\frac{1}{t}$ approaches $0$ from the left; and conversely, as $\frac{1}{t}=x$ approaches $0$ from the left, the value of $t=\frac{1}{x}$ approaches $-\infty$. So both limits are considering arguments that approach the same thing.

For (b), you forgot to change the limit as well: your limit is still a limit of $x$, not a limit of $t$. Notice that in (a), both the function and the argument (the stuff under the "$\lim$") change. So rather than $$\lim_{x\to 0^-}e^t$$ you should have a limit as $t\to-\infty$: $$\lim_{x\to 0^-}e^{\frac{1}{x}} = \lim_{t\to-\infty}e^{t}.$$ As to evaluating this limit, it should follow from what you know about the exponential function. This is a standard limit for the exponential.

For (c), use "a similar technique": set $t = \frac{1}{x^3}$. As $x\to\infty$, what happens to $t$? It approaches $0$. From what side? From the positive side. So $$\lim_{x\to\infty}\ln\left(\frac{2}{x^3}\right) = \lim_{x\to\infty}\ln\left(2\left(\frac{1}{x^3}\right)\right) = \lim_{t\to 0^+}\ln\left(2t\right).$$ Can you evaluate that limit?

Similarly for the limit as $x\to-\infty$. If $t=\frac{1}{x^3}$, what happens to $t$ as $x\to-\infty$?

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Absolute values are in order for the $x\to-\infty$ part of (c) to avoid evaluating $\ln$ at negative numbers. (Unless, part of the point of your leading questions is to realize this, in which case I apologize for spoiling the surprise!) –  wckronholm Jun 13 '11 at 20:47
Ok, I understand b now, I think. Because if I didn't change the limit, then asking how $x$ changes would have $e^t$ be a constant because there's nothing for $x$ to refer to...right? –  OghmaOsiris Jun 13 '11 at 22:23
@OghmaOsiris: If someone just came upon $\lim\limits_{x\to 0^-}e^{t}$ without any background, the answer they would give would most likely by "$e^t$, because there is no $x$"; which I think is what you are trying to say. But what you need to remember is that there was an $x$, "hidden" inside of $t$, since $t=\frac{1}{x}$ in that expression; but to make computation of the limit straightforward, you should change the limiting operation as well as the notation in the function. –  Arturo Magidin Jun 14 '11 at 3:46
Ok. That makes sense. Thanks so much! This makes much more sense to me now! –  OghmaOsiris Jun 14 '11 at 13:25

(a) $f(x)$ is the inverse of $t$ ... incorrect

(b) $x \to 0^+$ does not convert to $t \to 0^+$

(c) First do (a) and (b), then use the insight you gain to do (c). Note: you gain insight when you do them yourself, not when others do them for you...

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...when did I ask people to do them for me... I asked for help on how to do them. –  OghmaOsiris Jun 13 '11 at 20:03

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