L'hopitals Rule for Solving Limits with Complex Functions

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[SOLVED] L'hopitals Derivative

Homework Statement



lim x->0 \frac{1-x^2-e^{-x^2}}{x^4}I ended up using l'hopitals rule 4 times before I got the answer. And I got an answer of -1/2. I was wondering if someone could check that to see if its right. I would post all my work but that would take a while, however, if you want to see more of my steps ill be glad to do so. I tried graphing it on my calculator, but depending where I put the parenthesis it was right or wrong. Thanks
 
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My final equation looked like so...

\frac{16x^4e^{-x^2} - 24x^2e^{-x^2} + 16x^2e^{-x^2} - 8e^{-x^2} + 8x^2e^{-x^2} - 4e^{-x^2}}{24}
 
You can start by replacing x^2 with y in the beginning for nicer equations. I have not been doing anything with l'hopital yet, but I used the infinite series of e^x and got the same answer as you.
 
Ok, I have another question. The derivative of -e^{-x^2} would be 2xe^{-x^2} is that correct?
 
i just replaced it with y and that was much easier thanks
 
yes the derivative is correct as is your solution.
 
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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