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Need help checking my working with this limit at infinity.

$$\lim_{x\to \infty} (1+x^2)^{1/\ln x}$$

All help is appreciated

My working was-

$$\ln(1+x^2)^{1/\ln(x)} = \frac {\ln(1+x^2)}{\ln(x)}$$

Using L'Hopital

$$\frac {\frac {2x}{1+x^2}}{\frac {1}{x}}$$

$$\frac {2x}{1+x^2}$$

My question is does the 1 here simply get cancelled as x approaches infinity. That would give me the correct answer but I want to know that I didn't accidentally get to it in which case my working would be wrong. $$2=\lim_{x\to \infty} \ln(1+x^2)^{1/\ln(x)}$$

$$e^2=\lim_{x\to \infty} (1+x^2)^{1/\ln(x)}$$

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    $\begingroup$ You get ${2x^2\over 1+x^2}$ after using L'Hopital. $\endgroup$ – Edu Aug 7 '17 at 13:19
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    $\begingroup$ what is $n$ here in your formula? $\endgroup$ – Dr. Sonnhard Graubner Aug 7 '17 at 13:20
  • $\begingroup$ So you really mean $\ln\Big((1+x^2)^{1/\ln(x)}\Big)$ and not $\Big(\ln(1+x^2)\Big)^{1/\ln(x)}$ right ? $\endgroup$ – Surb Aug 7 '17 at 13:29
  • $\begingroup$ Please get rid of $n$ in your question. $\endgroup$ – zhw. Aug 7 '17 at 15:23
  • $\begingroup$ @zhw. sorry. Thank you for pointing that out $\endgroup$ – Virens Aug 7 '17 at 18:08
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Simplifying the fraction after applying L'Hopital, you should get $$\frac{2x^2}{1+x^2}$$ instead of $$\frac{2x}{1+x^2}$$

Then to get $$\lim_{x \to \infty} \frac{2x^2}{1+x^2}$$ you can again apply L'Hopital and get $$\lim_{x \to \infty} \frac{2x^2}{1+x^2}=\lim_{x \to \infty} \frac{4x}{2x}=2$$

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