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

Question

25377 views around the world

You can reuse this answer
Creative Commons License

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.