How do you find the limit of #(3x+1/x) - (1/sinx)# as x approaches 0 using l'hospital's rule?

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Answer 1 · 2016-11-09T20:51:23

You can't evaluate the last expression using L'Hôpital's rule as it is not of an indeterminate form #0/0# or #oo/oo#

Answer 2 · 2016-11-09T21:49:32

This is a bit of a non-answer...

#(3x+1/x)-(1/sin x) = (3x^2 sin x + sin x - x) / (x sin x)#

Let:

#{ (f(x) = 3x^2 sin x + sin x - x), (g(x) = x sin x) :}#

Let:

#I = (-1, 1)#

Then:

#f(x)# and #g(x)# are differentiable on #I#
#lim_(x->0) f(x) = lim_(x->0) g(x) = 0#
#g'(x) = sin x + x cos x != 0# when #x in I "\" {0}#
#lim_(x->0) (f'(x))/(g'(x)) = lim_(x->0) (6x sin x + 3x^2 cos x + cos x - 1)/(sin x + x cos x)#

which is again in the form #0/0#

So it's not clear that #lim_(x->0) (f'(x))/(g'(x))# exists and the new form is more complicated than the original problem.

So L'Hôpital's rule does not help us much.