How do you find the limit of #sin^2x/x# as x approaches 0?

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Answer 1 · 2016-01-29T06:04:03

#0#

Know the following limit identity:

#lim_(xrarr0)sinx/x=1#

We can rewrite the given function so that we can make use of the fact that #lim_(xrarr0)sinx/x=1#.

The question rewritten is

#lim_(xrarr0)sin^2x/x#

Notice that we can isolate #sinx/x# from this.

#=lim_(xrarr0)sinx/x(sinx)#

Limits can be multiplied, as follows:

#=lim_(xrarr0)sinx/x*lim_(xrarr0)sinx#

Since the first part equals just #1#, this simplifies to be

#=lim_(xrarr0)sinx#

We can now evaluate the limit by plugging in #0# for #x#.

#=sin(0)=0#

The function should approach #0# at #x=0:#

graph{(sinx)^2/x [-6.243, 6.243, -1, 1]}