Question #51d9b

1 Answer
Καδήρ Κ.
Dec 26, 2017

#lim_{x->0^+}(sin2x)^(x^(-2))=0#

Explanation:

#lim_{x->0^+}(sin2x)^(x^(-2))=lim_{x->0^+}(sin2x)^(1/x^2)=#

#lim_{x->0^+}e^(1/x^2*ln(sin2x))#

Let's calculate this limit now :

#lim_{x->0^+}ln(sin2x)*1/x^2=-oo#

So now our limit becomes:

#lim_{u->-oo}e^u=0#

Related questions
See all questions in Limits Involving Trigonometric Functions
Impact of this question
1692 views around the world
You can reuse this answer
Creative Commons License