Limit with trigonometric functions

Jules18
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Homework Statement



True or false?

lim Tanx/(1-Cosx) = lim Sec2x/Sinx = Infinity

(limits are as x approaches \pi from the left)

The Attempt at a Solution



I tried just plugging \pi into x in the first limit, and I ended up getting 0/2, which exists but is just 0. So I said false. Am I simplifying it too much? because what I did seems to make sense but it looks like that's not what they wanted me to do.
 
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Isn't it just 0/2=0?
 


The original limit was neither of the indeterminate forms [0/0] or [infinity/infinity], so L'Hopital's Rule doesn't apply here. Your first approach was correct.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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