Stuck on lim[x->0+] (ln x)^3 (arctan(ln(x+x^2))+pi/2) + (ln x)^2

[syndic8]

(ln x)³ (arctan(ln(x+x²))+pi/2) + (ln x)²


[Edit: Limit as x tends to 0+]

Trying to figure this out. I've plugged it into WolframAlpha, MATLAB and the answer is 1/3, but I need to show working by hand. I have tried to factor out (ln x)³ from the whole equation to render it into an indeterminate form (0/0) to apply L'Hopital's rule. This looks promising as I can factor 1/3 out of the limit, but I'm a bit stuck otherwise (further manipulation seems to be going nowhere). I've also tried series expansion of arctan, but it seems to be counter-productive to trying to render an indeterminate form or cancelling any terms.

 

[ksdhart2]

You forgot to include what value x is approaching! Without that bit of information the problem is impossible to solve. For instance:

\(\displaystyle \displaystyle \lim_{x \to 1} \left[ ln(x)^3 \cdot \left( tan^{-1} (ln(x+x^2))+\frac{\pi}{2} \right) + ln(x)^2 \right] = 0 \)

\(\displaystyle \displaystyle \lim_{x \to 2} \left[ ln(x)^3 \cdot \left( tan^{-1} (ln(x+x^2))+\frac{\pi}{2} \right) + ln(x)^2 \right] = ln(2)^3 \cdot \left( tan^{-1} (ln(2+2^2))+\frac{\pi}{2} \right) + ln(2)^2 \approx 0.333 \cdot 2.633 + 0.48 \approx 1.357\)

\(\displaystyle \displaystyle \lim_{x \to 57} \left[ ln(x)^3 \cdot \left( tan^{-1} (ln(x+x^2))+\frac{\pi}{2} \right) + ln(x)^2 \right] = ln(57)^3 \cdot \left( tan^{-1} (ln(57+57^2))+\frac{\pi}{2} \right) + ln(57)^2 \approx 66.089 \cdot 3.019 + 16.346 \approx 215.869\)

\(\displaystyle \displaystyle \lim_{x \to \infty} \left[ ln(x)^3 \cdot \left( tan^{-1} (ln(x+x^2))+\frac{\pi}{2} \right) + ln(x)^2 \right] = \infty\)

I cannot possibly guess which of the literally infinite number values x could be approaching would lead to a limit of 1/3.

 

[stapel]

(ln x)³ (arctan(ln(x+x²))+pi/2) + (ln x)²

[Edit: Limit as x tends to 0+]

For future reference, it is generally better to "reply" with new information, rather than implanting it (via "edit") in previous posts. If I hadn't happened to view this thread, it's possible that nobody would have noticed that there's now new information.

Trying to figure this out. I've plugged it into WolframAlpha, MATLAB and the answer is 1/3, but I need to show working by hand. I have tried to factor out (ln x)³ from the whole equation to render it into an indeterminate form (0/0) to apply L'Hopital's rule. This looks promising as I can factor 1/3 out of the limit, but I'm a bit stuck otherwise (further manipulation seems to be going nowhere). I've also tried series expansion of arctan, but it seems to be counter-productive to trying to render an indeterminate form or cancelling any terms.

It might help if you showed what you've tried. It might be that you were on the right track, and made a slightly-wrong turn right at the end. For instance, how did you get to a stage where you were factoring out "1/3"?

Thank you!