Solving for m: Find Limit of [e^(mx^2)-cos(8x)]/x^2 = 64

  • Thread starter Thread starter sparkle123
  • Start date Start date
  • Tags Tags
    Limit
sparkle123
Messages
172
Reaction score
0
Find m:
lim([e^(mx^2)-cos(8x)]/x^2)=64 where x-->0

I used l'hopital's twice and got
lim [2m^2x^2e^(mx^2)+32cos(8x)]
=32

which means the limit is never 64.
the right answer is m=32 though. Where am I wrong?
 
Physics news on Phys.org
Redo the second derivative of your numerator. Especially the first term.
 
Got it! Thanks :)
 
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...
Back
Top