Can $\displaystyle \lim_{x\rightarrow 0} \left(\frac{1}{x^2}\right)^x$ be solved as $\displaystyle \lim_{x\rightarrow 0}\frac{1^x}{x^{2x}}$? It is in the indeterminate form of ${\infty}^0$ so I think I should transform it to the $\ln$ form and solve it. But I am not sure if I can solve it as given above.
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2The indeterminate form is $\infty^0$, and like all exponential indeterminate forms, a good approach is to take logarithms and use L'hôpital's rule. – Sammy Black Mar 03 '24 at 21:41
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@SammyBlack thank you for the correction. I read about it but I do not understand why I can not use the method that i followed? Thank you very much – Philomath Mar 03 '24 at 21:45
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@Philomath It looks like Sammy Black suggested the same thing that you were going to try - taking the logarithm. The only issue is that you'll need L'hopital's rule after taking the logarithm, since after taking the logarithm it will be of the form $\infty \cdot 0$. – Polygon Mar 03 '24 at 22:02
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Does this multiduplicate answer your question? What is the limit of $x^x$ as $x$ approaches $0$ ($\lim\limits_{x \to 0}x^x$)? Yes it does: it shows that $L:=\lim_{x\to0}x^x=1$, and your limit is just the variant $\frac1{L^2}$. – Anne Bauval Mar 05 '24 at 11:19
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Notice that
$$\left(\frac{1}{x^2}\right)^x=(x^{-2})^x=x^{-2x}=e^{-2x\ln x}$$
Now utilize that $\lim_{x \rightarrow 0^+} x \ln x = 0$ which results in the limit equaling $1$.
Tanamas
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In almost every standard textbook for 1st year calculus, there exists a technique for calculating a limit of function, that is, take the logarithm of that function, calculate the limit, then see what is the exponential of that limit.
So here, we take its logarithm, then we have
$$\ln (\frac{1}{x^2})^x$$
$$= x \ln\frac{1}{x^2}$$
$$= x \ln\frac{1}{x^2}$$
$$= x \ln x^{-2}$$
$$= -2x \ln x$$
And this means
$$\lim_{x \rightarrow 0} \ln(f(x)) = \lim_{x \rightarrow 0}(-2x \ln x)$$
Since we know that
$$\lim_{x \rightarrow 0} x \ln x = 0$$
(It is easy to show this through L'Hôpital's rule)
we have
$$\lim_{x \rightarrow 0} \ln(f(x)) = -2 \times 0 = 0$$
Therefore,
$$\lim_{x \rightarrow 0} f(x)$$
$$= \lim_{x \rightarrow 0} e^{\ln f(x)}$$
$$= e^{\lim_{x \rightarrow 0} \ln f(x)}$$
$$= e^0$$
$$= 1$$
Hope this can help you.
ZYX
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