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I have to find the value of the function $\frac{\log(1+x^4)}{x^2 \tan^2(2x)}$ for $x$ that goes to zero.

It's an undefined form $\frac{0}{0}$ so I transformed the function in this way in order to use two known limits

$$\frac{\log(1+x^4)}{x^2} \frac{x^2}{x^2} \frac{1}{\tan^2(2x)} \frac{4x^2}{4x^2} $$ and so $$\frac{\log(1+x^4)}{x^4}\frac{4x^2}{\tan^2(2x)}\frac{x^2}{4x^2} $$

It seems to me that the value of the limit in zero is $\frac{1}{4}$ but looking to the graph of the function (using Desmos) it do not have a clear behavior in zero.

Can you tell me if my reasoning is right or wrong and, if wrong, where is the mistake?

Thank you.

ps. I found the problem with Desmos. In my notation I was using log for the logarithm with e as the base. But I should use ln in Desmos for the same logarithm, which otherwise is 10 based.

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    $\begingroup$ Your argument looks fine to me. Computer programs are not always reliable. $\endgroup$
    – Gary
    May 11, 2023 at 12:54
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    $\begingroup$ Something's off with the Desmos graph : it really should show a limit of $\frac 14$, but instead it shows something like $0.11$ or something, which is miles off. Wolfram Alpha does a better job of guessing the answer (without needing to zoom in). $\endgroup$ May 11, 2023 at 12:58
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    $\begingroup$ The way you have used is very fine, by standard limits indeed $$\frac{\log(1+x^4)}{x^2 \tan^2(2x)}=\frac14\frac{\log(1+x^4)}{x^4}\frac{4x^2}{\tan^2(2x)}\to \frac14 \cdot 1\cdot 1=\frac14$$ $\endgroup$
    – user
    May 11, 2023 at 13:03

2 Answers 2

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A systematic method is to use asymptotic development : All development are undermeant to be done at $x=0$.

$$ \log(1+x^4)=x^4 + o(x^4) \sim x^4$$ $$\tan^2(2x)=(2x +o(x))^2 = (4x^2+o(x)) \sim 4x^2$$

Here by the use of only multiplicative operation on equivalent :

$$\dfrac{\log(1+x^4)}{x^2\tan^2(2x)} \sim \dfrac{x^4}{x^2\times 4x^2}\sim\dfrac{1}{4} $$

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To solve the Limit, I have used a method of series expansion. I have attached an image, kindly go through the solution.

enter image description here

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  • $\begingroup$ Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. $\endgroup$ May 11, 2023 at 18:53

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