Limit question! Will be posted below.
\[\lim_{x \rightarrow 0^+} 3\sqrt{x}lnx\]
have you tried writing it as a quotient and see if l'hospital applies
I tried getting the derivative of it. But the answer key says the answer 0, and I'm not entirely sure why. @freckles
Getting the derivative and applying l'hospital's*
let's just ignore the 3 part for now
\[\lim_{x \rightarrow 0^+} \frac{\ln(x)}{\frac{1}{x^\frac{1}{2}}}\]
here we have -inf/inf so we can use l'hospital
the derivative of ln(x) w.r.t. x is ? the derivative of 1/(x^(1/2)) is w.r.t. x is?
1/x and -0.5sqrtx ?
\[\frac{d}{dx}(\frac{1}{x^{\frac{1}{2}}})=\frac{d}{dx}(x^{-\frac{1}{2}})=-\frac{1}{2}x^{\frac{-1}{2}-1}=\frac{-1}{2}x^{\frac{-3}{2}}=\frac{-1}{2 x^{\frac{3}{2}}}\]
\[l im_{x \rightarrow 0^+}\frac{\frac{1}{x}}{\frac{-1}{2 x^{\frac{3}{2}}}} \\ =\lim_{x \rightarrow 0^+} \frac{1}{x} \cdot \frac{2x^{\frac{3}{2}}}{-1} \]
Oh my god, careless mistakes. I get it now. Thank you! @freckles
np
Join our real-time social learning platform and learn together with your friends!