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

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Answer 1 · 2016-08-04T12:19:47

-1

#color(blue)(1^(st) " Method")#

First we check if we can use L'Hôpital's rule. For this to be viable we need the limit to be of indeterminate form.

#(sin(0))/(2(0)^2 - 0) = 0/0#

So limit is indeterminate and we can use L'Hôpital's.

#lim_(xrarr0)(sinx)/(2x^2-x) = lim_(xrarr0)(d/(dx)(sinx))/(d/(dx)(2x^2-x))#

#=lim_(xrarr0)(cosx)/(4x-1) = (cos(0))/(4(0)-1) = 1/(-1) = -1#

#color(blue)(2^(nd) " Method")#

Rewrite limit as

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

It's a logical fallacy to use L'Hôpital's on the first term so we use alternative methods such as the Squeeze Theorem to show that

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

So overall limit becomes:

#lim_(xrarr0)(sinx)/(2x^2-x) = lim_(xrarr0)(1)/(2x-1) = 1/(2(0) - 1) = 1/(-1) = -1#