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Please help me find $$\lim_{x\rightarrow \infty} \left(x -\ln({x^2}+1)\right)$$

It seems a tip were to factorize with $x^2$

Philippe

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    Hint: replace $x$ with $\ln{e^x}$ then combine the logarithms to obtain a fraction. Then apply L'Hopital's rule to the fraction. – 2017-01-10

4 Answers 4

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$$\lim _{ x\rightarrow +\infty } \left( x-\ln { \left( 1+x^{ 2 } \right) } \right) =\lim _{ x\rightarrow +\infty } \ln { \left( \frac { e^{ x } }{ 1+x^{ 2 } } \right) } \overset { L'hospital }{ = } \lim _{ x\rightarrow +\infty } \ln { \left( \frac { e^{ x } }{ 2x } \right) } \overset { L'hospital }{ = }$$$$= \lim _{ x\rightarrow +\infty } \ln { \left( \frac { e^{ x } }{ 2 } \right) } =+\infty $$

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The answer is infinity. $$\lim_{x\rightarrow \infty }\left ( x-\ln\left ( 1+x^{2} \right ) \right )=\lim_{x\rightarrow \infty }\ln\left ( \frac{e^{x}}{1+x^{2}} \right )=\ln\lim_{x\rightarrow \infty }\left ( \frac{e^{x}}{1+x^{2}} \right )\overset{[1]}{=}\infty$$

[1]use L'Hopital twice.

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    The second step is wrong (strictly speaking, $\frac{e^{x}}{1+x^{2}}\neq\frac{e^{x}}{x^{2}}$). The last step requires an explanation of some sort. – 2017-01-10
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    @barak manos when $x$ goes to infinity, can't we say that $1+x^{2}\sim x^{2}$? – 2017-01-10
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    You're not working on $x^2+1$, you're working on $\frac{e^x}{x^2+1}$. – 2017-01-10
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$x + y = x(1+\frac{y}{x})$

So, $\left (x- ln(x^{2} + 1) \right ) = \left ( x\left ( 1-\frac{ln(x^{2} + 1}{x} \right ) \right )$

$\lim_{x\rightarrow \infty } \left ( A(x).B(x) \right ) = \infty$

When $\lim_{x\rightarrow \infty }\left ( A(x) \right ) = \infty$ and $\lim_{x\rightarrow \infty }\left ( B(x) \right ) != 0$

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$$\lim_{x\to +\infty }\dfrac{\ln({x^2}+1)}{x}\underbrace{=}_{\text{L'Hopital}}0\Rightarrow\lim\limits_{x\rightarrow +\infty} \left(x -\ln({x^2}+1)\right)$$ $$=\lim\limits_{x\rightarrow +\infty} x\left({1 -\dfrac{\ln({x^2}+1)}{x}}\right)=+\infty.$$