How do you evaluate the limit of #lim (xcosx)/(x^2+x)# as #x->0#?

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Answer 1 · 2017-05-01T18:51:08

#lim_(x->0)(xcosx)/(x^2+x)=1#

We are trying to find #lim_(x->0)(x cosx)/(x^2+x)#. In order to do this, we need to apply L'hopital's rule:

#lim_(x->0)(x cosx)/(x^2+x)=lim_(x->0)(d/dx(xcosx))/(d/dx(x^2+x)#

#=lim_(x->0)(cosx-xsinx)/(2x+1)#

#=(cos0-0xxsin0)/(2xx0+1)#

#=1/1#

#=1#

Answer 2 · 2017-05-01T21:57:25

#lim_(x rarr 0)=1#

We can start by factoring the denominator.

#lim_(x rarr 0)=(xcos(x))/(x(x+1))#

We can cancel the #x# factor.

#lim_(x rarr 0)=(cos(x))/(x+1)#

Now evaluate the limit

#lim_(x rarr 0)=(cos(0))/(0+1)=1/1=1#