Question

Determine the limit as x goes to 1 of (x/x-1)∫ from 1 to x of [(arcsin t)/t]dt Can anyone help with this?

Answer 1

can you draw the original question

Answer 2

i can try

Answer 3

thx

Answer 4

sorry for the wait
here I wrote it in msword http://snag.gy/uLVu0.jpg

Answer 5

did you try plugging in x = 1 , into that integral, i think it produces zero

Answer 6

woops

Answer 7

ok lets do Lhopital

Answer 8

@perl \(\large\displaystyle \lim_{x\to 1}\frac{x}{x-1}\) doesn't exist since
\[\large \lim_{x\to 1^-}\frac{x}{x-1} = -\infty \neq \lim_{x\to 1^+}\frac{x}{x-1}=\infty\].
To evaluate this limit, you will need to use L'Hopital's rule since
\[\large \lim_{x\to 1} \frac{x{\displaystyle \int_1^x\frac{\arcsin t}{t}\,dt }}{x-1}\rightarrow\frac{0}{0}\] Hence,
\[\large \lim_{x\to 1} \frac{x{\displaystyle \int_1^x\frac{\arcsin t}{t}\,dt }}{x-1} = \lim_{x\to 1} \frac{\dfrac{d}{dx}\left(x{\displaystyle \int_1^x\frac{\arcsin t}{t}\,dt }\right)}{\dfrac{d}{dx}\left(x-1\right)}=\ldots\]