How do you find ? lim_(x->0)int_(0)^(2x)(e^(t^2))/(x+tan(x))

1 Answer
Oct 19, 2017

Look below.

Explanation:

lot of people would just evaluate int_0^{2x} e^{t^2}/{x+tan(x)}

but we have to find the limit, so do this

lim_{x->0} int_0^{2(0)} e^{0^2)/{0+tan(0)}

you end up with 0/0

now use l'hopitals rule by do the derivative

lim_{x->0} d/dx int_0^{2x} {d/dx e^{x^2}}/{d/dxx+tan(x)

lim_{x->0} e^{2x}/{1+sec^2(x)

lim_{x->0} e^{2(0)}/{1+sec^2(0)}

= e^0/{1+1}

=1/2