What is the limit of $sin(2x)/x^2$ as x approaches 0?

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Answer 1 · 2014-12-11T19:33:37

$lim_{x to 0}{sin(2x)}/{x^2}$

by l'H $hat{"o"}$ pital's Rule (0/0),

$=lim_{x to 0}{2cos(2x)}/{2x}=lim_{x to 0}{cos(2x)}/{x}$

Since

${(lim_{x to 0^-}{cos(2x)}/x=1/0^{-} = -infty),(lim_{x to 0^{+}}{cos(2x)}/x=1/0^{+}=+infty):}$ ,

$lim_{x to 0}{cos(2x)}/x$ does not exist,

which means that

$lim_{x to 0}{sin(2x)}/{x^2}$ does not exist.

Let us look at the graph of $y={sin(2x)}/x^2$ .

enter image source here

The graph above indeed agrees with our conclusion.

I hope that this was helpful.

What is the limit of sin(2x)/x^2sin(2x)/x^2 as x approaches 0?