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$$\lim_{{x \to 0}} \frac{{2x \ln(1 + x) + x^3}}{{\sqrt{1 - e^{-(x^{4})}}}}$$

I want to find easy solution of this limit, because I know that it's possible to make some uses of L'Hôpital's rule and find the limit (which will be $2$), but it would cost a lot of time. But I wonder if there is easy solution (e.g making some changes to task and then easily find limit by L'Hôpital's rule).

I've tried to put $x$ and $x^3$ out of brackets, made some transformations of $\sqrt{1 - e^{-(x^{4})}}$, and some other attempts, but nothing worked out.

Any help would be much appreciated.

user21820
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Donald
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2 Answers2

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One can write down the following power series expansion near $x = 0$ (i.e., Maclaurin series): \begin{align*} \ln(1 + x) & = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots \\ e^{-x^4} & = \sum_{n = 0}^\infty \frac{(-1)^nx^{4n}}{n!} = 1 - x^4 + \frac{x^8}{2!} - \frac{x^{12}}{3!} + \dots. \end{align*} It follows that \begin{align*} \lim_{x\to 0} \frac{2x\ln(1 + x) + x^3}{\sqrt{1 - e^{-x^4}}} & = \lim_{x\to 0} \frac{2x\left(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots\right) + x^3}{\sqrt{x^4 - \frac{x^8}{2!} + \frac{x^{12}}{6!} - \dots}} \\ & = \lim_{x\to 0} \frac{2x\left(x + \frac{x^3}{3} - \frac{x^4}{4} + \dots\right)}{x^2\sqrt{1 - \frac{x^4}{2!} + \frac{x^8}{3!} - \dots}}. \end{align*} Can you finish the problem from here?

Chee Han
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Recall that $e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}$. Then, $$\frac{2x\ln(1+x) + x^3}{\sqrt{1 - e^{-x^4}}} = \frac{2x\ln(1+x) + x^3}{\sqrt{1 - \sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{n!}}} = \frac{2x\ln(1+x) + x^3}{\sqrt{ \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{4n}}{n!}}} = \frac{2x\ln(1+x) + x^3}{x^2\sqrt{ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{4n}}{(n+1)!}}}.$$

Note that $\lim_{x\to 0} \sqrt{ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{4n}}{(n+1)!}} = 1$. Thus, we can write $$\lim_{x\to 0}\frac{2x\ln(1+x) + x^3}{\sqrt{1 - e^{-x^4}}} = \lim_{x\to 0}\frac{2x\ln(1+x) + x^3}{x^2} = \lim_{x\to 0} 2\frac{\ln(1+x)}{x} + x.$$

Now, we can use L'Hôpital's rule to get $$\lim_{x\to 0}\frac{2x\ln(1+x) + x^3}{\sqrt{1 - e^{-x^4}}} =\lim_{x\to 0} 2\frac{\frac{1}{1+x}}{1} + x = 2$$.

Or did I miss something?

Kuzja
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  • I don't know if after noting that $\lim_{x\to 0} \sqrt{ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{4n}}{(n+1)!}} = 1$ we can just put $1$ in the problem. But if it's true (i hope someone will confirm that), you have a good solution! – Donald Mar 10 '24 at 20:19
  • If you have $\lim_{n\to\infty} a_n = a$ and $\lim_{n\to\infty} b_n = b$, and $a,b \in \mathbb{R}$, then $\lim_{n\to\infty} a_n \cdot b_n= a \cdot b$. – Kuzja Mar 10 '24 at 20:22
  • OK it's true but why does $\lim_{x\to 0} \sqrt{ \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{4n}}{(n+1)!}} = 1$? Why $\sum_{n=0}^{\infty}(-1)^{n} = 1?$ – Donald Mar 10 '24 at 20:27
  • Oh, no. The first element in the series (for n = 0) is 1. All other elements are converging to zero. Note also that 0! = 1. – Kuzja Mar 10 '24 at 20:29
  • I think I understood. Thank you very much for your answers. – Donald Mar 10 '24 at 20:31