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Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$

I can otherwise solve this limit.This is in the form of $1^{\infty}$.

$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}}=\exp \left[{\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}-1\right)\times \frac{3x^2+8}{x^2}}\right]=e^{-8}$$ But i want to know,cant it be found using Taylor/Maclaurin series?What are the limitations of Taylor/Maclaurin series?
When can we not use them?
Is it true that we can use them whenever L Hospital rule is applicable(in $\frac{0}{0}$ and $\frac{\infty}{\infty}$ cases.)Can we use them in $0^0,\infty^0,1^\infty$ cases.
Please help me.Thanks.

leonbloy
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Vinod Kumar Punia
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4 Answers4

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Using logarithms and Taylor expansions can help a lot. $$A=\Big(\frac{3x^2+2}{5x^2+2}\Big)^{\frac{3x^2+8}{x^2}}$$ $$\log(A)=\frac{3x^2+8}{x^2}\,\log\Big(\frac{3x^2+2}{5x^2+2}\Big)=\frac{3x^2+8}{x^2}\,\log\Big(1-\frac{2x^2}{5x^2+2}\Big)\approx -\frac{3x^2+8}{x^2}\times\frac{2x^2}{5x^2+2}$$

I am sure that you can take from here.

Claude Leibovici
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Start by reducing to "known" Taylor series around $0$, and then compose them. Here, you will rely on that of $\ln(1+x)$, $\frac{1}{1+x}$, and the fact that $a^b = e^{b\ln a}$: $$ \left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}} = \left(\frac{1+\frac{3x^2}{2}}{1+\frac{5x^2}{2}}\right)^{\frac{3x^2+8}{x^2}} = e^{\frac{3x^2+8}{x^2}\ln\frac{1+\frac{3x^2}{2}}{1+\frac{5x^2}{2}}} $$ Now, $$ \frac{1+\frac{3x^2}{2}}{1+\frac{5x^2}{2}} = 1-x^2 +o(x^3) $$ (using the known Taylor series expansion of $\frac{1}{1+x}$), and as $\ln(1+x)= x + o(x)$ around $0$ you get $$ \left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}} = e^{\frac{3x^2+8}{x^2}(-x^2 +o(x^2))} = e^{-8 + o(1)} $$

To answer the more general question: in my experience, you can use Taylor series for limits around $0$, $\pm\infty$ (after some factorization to reduce to a limit around $0$), etc., whenever L'Hospital is applicable. (And I would argue in favor of the former over the latter almost every time, maybe out of personal taste.) Moreover, once you get to memorize the usual suspects (standard Taylor series of common functions), then it becomes quite fast (albeit sometimes cumbersome). The main advantage being: it is not always the most elegant method, but it is systematic -- if you don't see an other obvious way to compute the limit, try the Taylor expansions.

Clement C.
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$t=e^{\log t}$. Now do some algebra with the logged expression to get something to apply Maclaurin series.

Alex
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Notice, we know $\color{blue}{\lim_{n\to \infty}\left(1-\frac{1}{n}\right)^n=e^{-1}}$

Now we have

$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$$$=\lim_{x\to 0}\left(1-\frac{2x^2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$ let, $\large \frac{2x^2}{5x^2+2}=\frac{1}{t}$$\implies x^2=\large \frac{2}{2t-5}$, hence $$=\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^{8t-17}$$ $$=\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^{8t}\left(1-\frac{1}{t}\right)^{-17}$$ $$=\left(\lim_{t\to \infty}\left(1-\frac{1}{t}\right)^t\right)^{8}\cdot \lim_{t \to \infty}\left(1-\frac{1}{t}\right)^{-17}$$ $$=(e^{-1})^{8}\cdot (1)^{-17}=\color{red}{\large e^{-8}}$$

Harish Chandra Rajpoot
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  • Just to check something: at first glance, it seems that in your substitution (after "let,"), $t$ has to be negative (since the LHS is positive). Shouldn't $t$ then go to $-\infty$, not $\infty$? Or the signs somewhere else reflect this? – Clement C. Dec 05 '15 at 14:53
  • @Clement C: Thanks for your observation. I fixed the error – Harish Chandra Rajpoot Dec 05 '15 at 14:59
  • You're welcome... this should be the reason why you find $e^8$ (instead of $e^{-8}$). – Clement C. Dec 05 '15 at 15:01