How do you find the limit of #sinx/(2x^2-x)# as x approaches 0?

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Answer 1 · 2016-01-25T19:49:29

#-1#

One thing you should know going into this is that

#lim_(xrarr0)sinx/x=1#

This is an important identity in limit problems.

Keeping this in mind, we can factor an #x# from the denominator of the fraction, giving

#=lim_(xrarr0)(sinx)/(x(2x-1)#

We can rearrange this to get #sinx/x#, which we already know the limit of.

#=lim_(xrarr0)(sinx/x)1/(2x-1)#

Limits can be multiplied, as follows:

#=lim_(xrarr0)sinx/x*lim_(xrarr0)1/(2x-1)#

Since #lim_(xrarr0)sinx/x=1#, the expression simply equals

#=lim_(xrarr0)1/(2x-1)#

And here, we can evaluate the limit straightaway by plugging in #0# for #x#.

#=1/(2(0)-1)=-1#

We can check the graph at #x=0#:

graph{sinx/(2x^2-x) [-5.206, 5.89, -2.9, 2.65]}