Question

Limit question:

Answer 1

\[\Large \lim_{x \rightarrow 0} \frac{1}{x^{3}} \int\limits_{0}^{x} \ln(1-t^{2})dt\]

Answer 2

i tried evaluating the limit hoping that something in the evaluation would cancel out the x^3 in the denominator but i had no such luck...any advice?

Answer 3

@amistre64 @experimentX

Answer 4

Looks like an Indeterminant Form, to me. Type 0/0.

Answer 5

yes, 0/0. just apply lhopital's rule

Answer 6

\[\Large \lim_{x \to 0} \frac{1}{x^{3}} \int\limits_{0}^{x} \ln(1-t^{2})dt\]
\[\Large \frac{\frac d{dx}\int\limits_{0}^{x} \ln(1-t^{2})dt}{\frac d{dx} x^3}\]
\[\Large \frac{ln(1-x^2)}{3x^2}\]
\[\Large \frac{\frac d{dx}ln(1-x^2)}{\frac d{dx}3x^2}\]
\[\Large \frac{\frac{-2x}{1-x^2}}{6x}\]
\[\Large \frac{-1}{3(1-x^2)}\]
now as x goes to zero?

Answer 7

the key is to recognize the LHop; and also what it means to take the derivative of an integration
\[\frac d{dx}~\int_{a(x)}^{b(x)}{f(t)~dt}\to ~\frac d{dx}~[F(b(x))-F(a(x))]\]
\[\frac d{dx}~[F(b(x))-F(a(x))]\to~f(b(x))b'-f(a(x))a'\]
in this case; f(t) = ln(1-t^2) therefore
\[\frac d{dx}~[F(b(x))-F(a(x))]\to~ln(1-x^2)-ln(1-0^2)\]

Answer 8

pfft, and i spose 0' = 0 so that would also just cancel the f(a) bits regardless :)

Answer 9

Thanks, that's very helpful. What would your final answer be for the question?

Answer 10

uhhh .... as long as i didnt mess up along the way; im sticking with my original solution
\[\Large \lim_{x\to0} ~\frac{-1}{3(1-x^2)}\]

Answer 11

the solution i have is 3/10

Answer 12

http://www.wolframalpha.com/input/?i=limit+%28x+to+0%29+ln%281-x%5E2%29%2F3x%5E2

Answer 13

the solution that you came up with? or is it a solution from a key?