Compute $\lim\limits_{x\to 0^+}(1+x^2)^{\frac{1}{x}}$?

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so I started the casual way:$$\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 0^+}(1+x^2)^{\frac{1}{x}}=\lim\limits_{x\to 0^+}e^{\ln \left((1+x^2)^{\frac{1}{x}}\right)}$$ How can I proceed?

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You are almost done:

$$\lim\limits_{x\to 0^+}\ln \left((1+x^2)^{\frac{1}{x}}\right)=\lim\limits_{x\to 0^+}\frac{\ln \left(1+x^2\right)}{x}=\lim\limits_{x\to 0^+}\frac{\ln \left(1+x^2\right)- \ln(1+0^2)}{x-0}$$

is the definition of the derivative of $\ln(1+x^2)$ at $x=0$. Just derivate $\ln(1+x^2)$.

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This is equivalent to $$\lim_{x\to0^+}\left((1+x^2)^{1/x^2}\right)^x$$ using the fact that $$\lim_{f(x)\to0^+}(1+f(x))^{1/f(x)}=e$$ This limit is then of the form $\lim_{x\to 0^+} (g(x))^x$ where $\lim_{x\to0^+}g(x)=e\gt0$ hence the limit is not indeterminate and can be given directly by $e^{0^+}=1$.

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You can use the binomial series:

$$ (1+x^2)^{\frac{1}{x}}=1+\frac{1}{x}x^2+\frac{1}{x}\left(\frac{1}{x}-1\right)\frac{(x^2)^2}{2!}+\frac{1}{x}\left(\frac{1}{x}-1\right)\left(\frac{1}{x}-2\right)\frac{(x^2)^3}{3!}+\cdots=\\ 1+x+\frac{1-x}{x^2}\frac{x^4}{2!}+\frac{(1-x)(1-2x)}{x^3}\frac{x^6}{3!}+...=\\ 1+x+\frac{x^2(1-x)}{2!}+\frac{x^3(1-x)(1-2x)}{3!}+\cdots\xrightarrow[x\to 0^+]{}1+0+0+0+\cdots=1. $$

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As usual, I would do it with equivalents:

It is enough to find the limit of the log: $$\ln\bigl(1+x^2\bigr)^{\!\frac 1x}=\frac 1x\ln(1+x^2)\sim_0=\frac 1x\cdot x^2=x\xrightarrow[x\to 0]{}0.$$