Can Anyone Help Solve This Limit Evaluation Problem?

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alephnought
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Would anybody solve this problem for me?

I've tried it for a long time, but don't seem to get the answer.
I don't think I can apply L'Hospital's rule because the numerator is not zero or indeterminate while the denominator goes to zero

lim x-> -inf ((1+ e^(1/x))/e^x)

ok, if I assume the numerator is 1 - e^... and try to solve, I am not able to get rid of the e^x term


thanks
 
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One of the very first things you should have learned about limits is that if the denominator of a fraction goes to 0 and the numerator does not, then the fraction does not have a limit!
 
I think the answer is +\infty because a number divided by 0 tends to infinity.
 
LinkMage said:
I think the answer is +\infty because a number divided by 0 tends to infinity.

Yes, the limit is "infinity" which is just a way of saying that the limit does not exist. It really bothers me to see "a number divided by 0 tends to infinity"! A number cannot be divided by 0 and there is no dividing by 0 in this problem because the denominator is never 0 for any value of 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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