Question

Evaluate limit...

Answer 1

know about L'Hopital's rule ?

Answer 2

yeah.. but how do u differentiate the numerator which is the integral term

Answer 3

\(\large \dfrac{d}{dx}\int\limits_a^bf(x)dx = [f(x)]^b_a\)

Answer 4

its like derivative and integral cancels out and we only need to put those 2 limits

Answer 5

If you haven't learned about Calculus symbols yet don't use them in your answer, just a tip

Answer 6

i am taking calculus thats why T^T

Answer 7

Ah okay, I am also in Calculus but I got help on OS for a problem and my teacher told me not to use symbols yet because we hadn't discussed them...

Answer 8

anw, i need to use chain rule for this right, i did 2x times e^sqrtx^2

Answer 9

why do u need any rule ?
sqrt x^2 is just x

Answer 10

infact |x| but since x->2 , we are sure that x>0
and so |x| = x

Answer 11

isit haha my bad :p omygosh i actually did subs y=x^2 so ya, the ans is 2xe^x then?

Answer 12

how 2x e^x ?
how is 2x there ?

Answer 13

i did chain rule, like dy/dx=2x , df/dx=df/dy times dy/dx

Answer 14

omg is that correct?

Answer 15

\(\huge [e^{\sqrt t}]_4^{x^2} = e^{\sqrt {x^2}}-e^{\sqrt4} = e^{|x|} - e^2 = e^x-e^2\)

Answer 16

you differentiated ?
d/dx and the integral actually cancelled....
have a look at the standard result i posted again

Answer 17

omg what am i doing haha, thats for the numerator?

Answer 18

yes, but did u get how ?
(thats after differentiating the numerator)

Answer 19

yeah i get it now (: so overall\[\lim_{x \rightarrow 2}\frac{ e^x-e^2 }{ 2x-2 }=0/2?\]

Answer 20

yes, i get 0 too

Answer 21

okay thank you! :D

Answer 22

wait, im kinda having doubts getting that answer, does it really cancel? since the integral is in dt right?

Answer 23

yes, i does get cancelled, but i think i missed something, lets do it by actually solving the integral
you got 2xe^x by solving the integral, and then putting the limits, right ?

Answer 24

really because the formula u gave me above, is in dx but int his case the integral is in dt

Answer 25

actually i meant to give this,
\(\large \dfrac{d}{dx}\int\limits_{f(x)}^{g(x)}h(z)dz = [h(z)]^{g(x)}_{f(x)}\)
but that become somewhat complicated
when we can easily integrate the given integral

Answer 26

using integration by parts

Answer 27

after substitution of sqrt t = u

Answer 28

okay so, t=u^2 then dt=2udu,

Answer 29

yes
integral 2u e^u du

Answer 30

okay \[f(x)=\int\limits_{4}^{x^2}e^\sqrt{t}dt=\int\limits_{2}^{x}2ue^udu\]

Answer 31

u need help integrating too ?
or working on it ?

Answer 32

i got \[\lim_{x \rightarrow 2}\frac{ 2xe ^{\sqrt{x^2}} }{ 2x-2 }=2e^2?\]

Answer 33

yes, that should be correct.
(we should have got 2e^2 using 1st method too, but idk where i was making error)

Answer 34

i see, hmm i have no idea, i thought it wasnt right to cancel straightaway because the integral and derivative is different., anyways thank you! :D

Answer 35

welcome ^_^