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