How do you find #lim (xlnx)/(x^2-1)# as #x->1# using l'Hospital's Rule?

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Answer 1 · 2016-12-13T20:44:37

#lim_(x->1) (xlnx)/(x^2-1) =1/2#

Let's put:

#f(x) = (xlnx)/(x^2-1) = f(x)/g(x)#

As:

#lim_(x->1) xlnx = 0#

and

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

we can use l'Hopital's rule stating that in such case:

#lim_(x->1) f(x)/g(x) = lim_(x->1) (f'(x))/(g'(x))#

if such limits exists.

#f'(x) = lnx+1#

#g'(x) = 2x#

#lim_(x->1) (f'(x))/(g'(x)) = lim_(x->1) (lnx+1)/(2x) =1/2#