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

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Answer 1 · 2017-10-19T13:04:51

Look below.

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#