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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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    @Arturo: Thanks again. I have saved your comments as "Limits per Arturo". We'll see if it helps my posts. I guess I like $\int\limits_a^b$ even though I am in the US.2011-02-16

2 Answers 2

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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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    @Arturo, if you use $t=\frac{s}{M}$ then you actually get $\frac{s}{t}\left(\frac{1-\exp(-t)}{s}\right)=\frac{1-\exp(-t)}{t}$.2011-07-16
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You could also use the power series expansion of the exponential function. More specifically, $e^x=1+x+O(x^2)$ as $x\to 0$.