<s>[INST] $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion 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.
 [/INST] 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.
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