Limit of f(x) as x tends to 0 for f(x)=[sinx]/[x^2]

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Homework Statement



Does f(x) tend to a limit as x tends to 0?

Homework Equations



f(x)=[sinx]/[x^2]

The Attempt at a Solution



Well i sinx would tend to zero and so would x^2, so would the limit just be zero?
 
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No, 0/0 is an indeterminant form. It means you have to work harder. Do you know the limit of sin(x)/x? Do you know l'Hopital's rule?
 
Do you know that:
\lim_{x \rightarrow 0}\frac{sinx}{x}=1

?
 
\frac{sin(x)}{x^2}= \left(\frac{sin(x)}{x}\right)\left(\frac{1}{x}\right)
 
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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