How Do Series Expansions Relate to Exponential Functions?

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limit as n→∞ of \frac{(2t)^n}{n!} and \frac{(-t)^n}{n!}

Answers are e2t-1 and e-t-1 but I don't know how to work them out, thanks.

edit: btw these are series
 
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If those are series you should fix up your question showing the limits of the series. Do you know ##e^a=\Sigma^\infty_0 \frac{a^n}{n!}##?
 
e^(i Pi)+1=0 said:
limit as n→∞ of \frac{(2t)^n}{n!} and \frac{(-t)^n}{n!}

Answers are e2t-1 and e-t-1 but I don't know how to work them out, thanks.

edit: btw these are series


Starting at n=1, it would seem? That would be unusual.
 
These are easy if you know that
\sum_{n=0}^\infty \frac{x^n}{n!}= e^x
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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