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I'm having some troubles computing this limit. $$\lim_{M\to\infty}M\left(\frac{1}{s} - \frac{e^{\frac{-s}{M}}}{s}\right)$$

I know that the answer should be 1, but I can't seem to figure out the steps to get there.

Here are the steps I've tried taking:

The exponential approaches 1 as $M\to\infty$, $$\lim_{M\to\infty}M\left(\frac{1}{s} - \frac{1}{s}\right)$$

Simplifying the statement leaves $\lim\limits_{M\to\infty}M * 0$

Where do I go from here, or have I made a mistake above?

As a side note, this is for deriving the laplace transform of the dirac-delta function by approximating the dirac-delta function as a finite rectangular pulse starting a t=0 and ending at $t=\frac{1}{M}$ with a magnitude of M, then taking the limit as $M\to\infty$

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You made a mistake. You cannot take bits and pieces of the limit and do them in whichever order you want; so you cannot replace $e^{-s/M}$ with its limit while still doing the limit. (If you could, you could argue that $\lim\limits_{x\to\infty}(x-x)$ is $\infty$, before the first $x$ goes to $\infty$, and $\infty-x=\infty$ for any real number $x$, or that it is $-\infty$ because the second $x$ goes to $\infty$, and $x-\infty=-\infty$ for all real numbers $x$). – Arturo Magidin Feb 15 '11 at 21:47
@Arturo: does \limits work for sums and integrals, too? I like to paste the problem into the title as it helps me (and I hope others) recognize the post. I had seen a claim that titles were always in display mode, but the $M \to \infty$ didn't come under the $\lim$ for me. Is this the answer? – Ross Millikan Feb 16 '11 at 4:25
@Ross: \limits works for operators (places limits directly above and below, as usual for, say, $\sum$; \nolimits places limits directly to the right, as is usual for integrals), but some symbols are not considered operators. If LaTeX complains, you can take the symbol and declare it an operator with \mathop{...}, and then \limits and \nolimits will work. \limits and \nolimits override whatever the "standard" is in whatever mode you're in. \sum\limits: $\sum\limits_a^b$; \sum\nolimits: $\sum\nolimits_a^b$; no \mathop needed here. – Arturo Magidin Feb 16 '11 at 4:29
@Arturo: Thanks. This is too complicated for me. But I guess there are so many ways people want to do things that it can't be simple... – Ross Millikan Feb 16 '11 at 4:38
@Ross: Just try \limits when you want them as we usually use them with $\lim$, $\sum$, and $\prod$ but they don't show up that way. If you get an error message about math operators, then enclose the symbol in \mathop, e.g., +_a^b: $+_a^b$; vs. +\limits_a^b: $+\limits_a^b$; vs. \mathop{+}\limits_a^b: $\mathop{+}\limits_a^b$. If the limits show up above and below and you want them in front, then use \nolimits, and if it complains you add the \mathop – Arturo Magidin Feb 16 '11 at 4:46
up vote 5 down vote accepted

Try rewriting the equation as $$\lim_{t\to 0^+}\left(\frac{1-e^{-t}}{0-(-t)}\right)$$ with the appropriate substitution. From there on it should probably be quite easy to figure out why the answer is 1.

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How did you convert a limit going to $\infty$ to one which approaches 0 (what's that rule called)? – helloworld922 Feb 15 '11 at 21:57
@helloworld922: He did a change of variable. For example, if you had a limit as $M\to 7$, and you let $t=2M+1$ to do a change of variable, then as $M\to 7$ you have that $t\to2(7)+1 = 15$. So the limit would change from being a limit as $M$ goes to $7$, to being a limit as $t$ goes to $15$. – Arturo Magidin Feb 15 '11 at 22:02
are you sure about your change of variable? And I think that limit should be one-sided. – Arturo Magidin Feb 15 '11 at 22:03
@arturo: I'm pretty sure i did it right, though it has been a while sinc calculus class, so i don't take responsibility for calculations ;). any way- i agree with you on the one sided limit thing. changed it – kneidell Feb 15 '11 at 22:15
(-: If I try $t=\frac{s}{M}$, I get $\frac{s-te^{-t}}{st}$, but you may have used a different one. – Arturo Magidin Feb 15 '11 at 22:20

You could also use the power series expansion of the exponential function. More specifically, $$e^x=1+x+O(x^2)$$ as $x\to 0$.

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