Question

lim = infinity 2+sinx / x^2 + 3​

Answer 1

Answer:

Given, x→0limx2sinxx−sinx=x→0limx3x−sinx.sinxx

=x→0limx3x−sinx.1  (∵x→0limxsinx=1)

Applying L' Hospital's Rule, we have

=x→0limx3x−sinx       ...00 form

=x→0lim3x21−cosx   ...00 form

=x→0lim6xsinx     ...00 form

=61x→0limxsinx=61.