Evaluation of a limit at infinity

Question

Evaluate $$\lim_{x \to +\infty} [\sqrt{x}-\ln(x^2+1)]$$

I tried to multiply both numerator and denominator by conjugate and tried applying L'hopital but the calculations become way too complex. Is there any clever way of doing it?

Answer 1

Intuitively, $\ln\left(x^2+1\right)$ goes to infinity like $\sqrt x$ as $x$ goes to infinity, but in a much slower way. You can show that $\lim_{x\to\infty}\ln\left(x^2+1\right)/\sqrt x=0$ for example using that for $x\geqslant 1$, $$ 0\leqslant \ln\left(x^2+1\right)\leqslant \ln\left(2x^2\right)=\ln 2+2\ln x $$ and that $\lim_{x\to\infty}\ln x/\sqrt x=0$.

Then conclude via $$ \sqrt x-\ln\left(x^2+1\right)=\sqrt{x}\left(1-\frac{\ln\left(x^2+1\right)}{\sqrt x}\right). $$

Answer 2

We have $$\lim_{x \to \infty} (\sqrt{x} - \log(x^2 + 1)) = \lim_{x \to \infty} \log(x^2 + 1)(\frac{\sqrt x}{\log(x^2 + 1)} - 1) \geq \\ \lim_{x \to \infty} \log(x^2 + 1)(\frac{\sqrt x}{3\log x} - 1) = \infty$$ since $\lim_{x \to \infty} \frac{x^r}{\log x} = \infty$ for $r > 0.$ I hope this helps. :)

Answer 3

The proof can be based only on the inequality $u\ge \log u.$ Indeed for $k>0$ we have $$\log (u^k)=k\log u\le ku$$

Next for $x\ge 1,$ $u=\sqrt[4]{x}$ and $k=8$ we obtain $$\sqrt{x}-\log(x^2+1)=u^2-\log (u^8+1) \\ \ge u^2-\log(2u^8)=u^2-\log( u^8)-\log 2\\ \ge u^2-8u-\log 2=u(u-8)-\log 2$$ Thus the limit is equal $\infty.$

Answer 4

Consider the exponential of our expression... $$ \exp[\sqrt{x}-\ln(x^2+1)] = \frac{e^\sqrt{x}}{x^2+1} $$ Then try substitution $\sqrt{x} = u$, $$ \lim_{u\to+\infty}\frac{e^u}{u^4+1} $$ Apply l'Hopital many times, as long as it has indeterminate form $\infty/\infty$... $$ \lim_{u\to+\infty}\frac{e^u}{4u^3} \\ \lim_{u\to+\infty}\frac{e^u}{12u^2} \\ \lim_{u\to+\infty}\frac{e^u}{24u} \\ \lim_{u\to+\infty}\frac{e^u}{24} = +\infty $$ Therefore $$ \lim_{x\to+\infty}\frac{e^\sqrt{x}}{x^2+1} = +\infty $$ and, finally, take the logarithm $$ \lim_{x\to+\infty} [\sqrt{x}-\ln(x^2+1)]= +\infty $$

Answer 5

We have that $\log(x^2+1)\sim\log(x^2)=2\log x$ and because $\lim_{x\rightarrow\infty}\frac{\log x}{x^n}$ is zero ($n>0$) the limit is just $\infty$.

Answer 6

Actually you can find a littl econclusion here,
In this kind of situation, you can neglect the influence from ln(f(x)), when we only consider the situation that f(x)=$x^k$
Here is a general view:
Suppose we are seeking for the value of $$\lim\limits_{ x\rightarrow\infty}[g(x)+\ln(f(x))]$$ Even suppose the power of $f(x)$ is astronaumically higher then $g(x)$.
You can always use the property of the logarism function to take the power out and change it into the coefficient infront of the ln, so we can actually deduce the power of logarism function to a very tiny amount.
Then in this question we will find that the limit value is only about the limit value of $\lim_{x\rightarrow \infty}\sqrt{x}$ which is always $\infty$