How to find limit(x->0)(1/x)integral(0 to x) 2010^(1-t^2) dt

Find the limit of
(1/x) * integral(0 to x) 2010^(1-t^2) dt

when x->0

In the integral the lower limit is 0 and the upper limit is x

lim(x→0) (1/x) ∫(t = 0 to x) 2010^(1 - t^2) dt
= lim(x→0) [∫(t = 0 to x) 2010^(1 - t^2) dt] / x; this is of the form 0/0
= lim(x→0) 2010^(1 - x^2) / 1, via L'Hopital's Rule and Fund Theorem of Calculus
= 2010^(1 - 0)
= 2010.

I hope this helps!