$\lim\limits_{x\rightarrow \infty} \left(x -\ln({x^2}+1)\right)$ 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
 A: $$\lim _{ x\rightarrow +\infty  } \left( x-\ln { \left( 1+x^{ 2 } \right)  }  \right) =\lim _{ x\rightarrow +\infty  } \ln { \left( \frac { e^{ x } }{ 1+x^{ 2 } }  \right)  } \overset { L'hospital }{ = } \lim _{ x\rightarrow +\infty  } \ln { \left( \frac { e^{ x } }{ 2x }  \right)  } \overset { L'hospital }{ = }$$$$= \lim _{ x\rightarrow +\infty  } \ln { \left( \frac { e^{ x } }{ 2 }  \right)  } =+\infty $$
A: 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.
A: $x + y = x(1+\frac{y}{x})$
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 $\lim_{x\rightarrow \infty }\left ( B(x) \right ) != 0$
A: $$\lim_{x\to +\infty }\dfrac{\ln({x^2}+1)}{x}\underbrace{=}_{\text{L'Hopital}}0\Rightarrow\lim\limits_{x\rightarrow +\infty} \left(x -\ln({x^2}+1)\right)$$ $$=\lim\limits_{x\rightarrow +\infty} x\left({1 -\dfrac{\ln({x^2}+1)}{x}}\right)=+\infty.$$
