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

2 Answers
Nov 9, 2016

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#

Nov 9, 2016

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

Explanation:

#(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.