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

Question

Emma Aldridge · Accepted Answer

#-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]}

William Abdalla · Answer

undefined To find the limit of sinx/(2x^2-x) as x approaches 0, we can use algebraic manipulation and the properties of limits. First, we can factor out an x from the denominator to get sinx/x(2x-1). Then, we can simplify sinx/x to 1, as sinx/x approaches 1 as x approaches 0. Now, we are left with the limit of (1)/(2x-1) as x approaches 0. Plugging in 0 for x gives us 1/(2(0)-1) = 1/(-1) = -1. Therefore, the limit of sinx/(2x^2-x) as x approaches 0 is -1.