Lim x to 0 (1/sinx - 1/x ) is equals to infinity?

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may i know is it lim x to 0 (1/sinx - 1/x ) is equals to infinity??
 
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You cannot apply L'Hospital on this function (yet), you'd need to rewrite it so that you get something of the form 0/0 or ∞/∞.
 
To apply L'Hospital's Rule, put the expression in theform f(x)/g(x). If the form is indeterminate, then you can apply the Rule.
 
daveb said:
To apply L'Hospital's Rule, put the expression in theform f(x)/g(x). If the form is indeterminate, then you can apply the Rule.
Just to avoid confusion: it has to be 0/0 or ∞/∞ as I said, since there are more indeterminate forms possible. Many can be reduced to one of the two 'allowed' ones to use the rule though.
 
okok.thanx
 

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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