Another way to compute $\lim\limits_{x\to+\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)$

Question

I need to compute as a title a limit with $x\to+\infty$. The only way I found to obtain a result is to use the L'Hôpital's rule:

$$\lim\limits_{x\to+\infty}x^2\log\bigg(\frac{x^2+1}{x^2+3}\bigg)=\lim\limits_{t\to 0}\frac{1}{t^2}\log\bigg(\frac{1+t^2}{1+3t^2}\bigg)\stackrel{H}{=}\lim\limits_{t\to 0}\frac{1}{2t}\bigg(\frac{2t}{1+t^2}-\frac{6t}{1+3t^2}\bigg)=\lim\limits_{t\to 0}\bigg(\frac{1}{1+t^2}-\frac{3}{1+3t^2}\bigg)=-2$$

It seems the result is the difference between the two functions inside the log, but how this can be possible? Is there another way to compute these type of limits? I mean a calculus way without using theorems.

Answer 1

You can write $$x^2 \ln \left( \frac{x^2+1}{x^2+3} \right) = x^2 \ln \left( 1- \frac{2}{x^2+3} \right) \sim x^2 \times \frac{-2}{x^2+3}$$

So the limit is $-2$.

Answer 2

Hint

Use $$\lim_{h\to0}\dfrac{\ln(1+h)}h=1$$

for $h=t^2,3t^2$ to get the limit $=1-3$

Answer 3

\begin{align}\lim_{x\to\infty}x^2\log\left(\frac{x^2+1}{x^2+3}\right)&=\lim_{x\to\infty}x\log\left(\frac{x+1}{x+3}\right)\\&=\lim_{x\to0}\frac{\log\left(\frac{x+1}{3x+1}\right)}x.\end{align}So, compute $f'(0)$, where $f(x)=\log\left(\frac{x+1}{3x+1}\right)$.

Answer 4

Let $z:=x^2+3$ and $$\lim_{z\to\infty}\left((z-3)\log\frac{z-2}{z}\right)=\lim_{z\to\infty}\left(z\log\left(1-\frac2{z}\right)\right) =\log\left(\lim_{z\to\infty}\left(1-\frac2{z}\right)^z\right) \\=\log e^{-2}.$$

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The term $-3$ can be ignored because the $\log$ vanishes.

Answer 5

Alternatively: $$\lim_{x\to+\infty} x^2 \ln \left(\frac{x^2+1}{x^2+3}\right)=\\ \lim_{x\to+\infty} \ln \left(\frac{x^2+1}{x^2+3}\right)^{x^2}=\\ \ln \left[\lim_{x\to+\infty} \left[\left(1+\frac{-2}{x^2+3}\right)^{\frac{x^2+3}{-2}}\right]^{\frac{-2x^2}{x^2+3}}\right]=\\ \ln e^{\lim_\limits{x\to+\infty} \frac{-2x^2}{x^2+3}}=\\ \ln e^{-2}=-2.$$

Answer 6

Write

$$x^2\log\left(x^2+1\over x^2+3\right)=x^2\int_{x^2+3}^{x^2+1}{dt\over t}=-x^2\int_{x^2+1}^{x^2+3}{dt\over t}$$

and note that, since $1/t$ is strictly decreasing for $t\gt0$, we have

$${(x^2+3)-(x^2+1)\over x^2+3}\lt \int_{x^2+1}^{x^2+3}{dt\over t}\lt {(x^2+3)-(x^2+1)\over x^2+1}$$

Simplifying the numerators and remembering to reverse the directions of inequalities when multiplying by a negative quantity (namely $-x^2$), we have

$${-2x^2\over x^2+3}\gt-x^2\int_{x^2+1}^{x^2+3}{dt\over t}\gt{-2x^2\over x^2+1}$$

and the Squeeze Theorem does the rest.

Answer 7

If it is OK for you to calculate directly a derivative then you may proceed as follows:

• Set $x^2 = \frac{1}{t}$ and consider $t\to 0^+$

So, you get

\begin{eqnarray*}x^2\log\bigg(\dfrac{x^2+1}{x^2+3}\bigg) & \stackrel{x^2 = \frac{1}{t}}{=} & \frac{\log \frac{1+t}{1+3t}}{t}\\ & = &\frac{\log (1+t) - 0}{t} - \frac{\log (1+3t) - 0}{t} \\ & \stackrel{t \to 0^+}{\longrightarrow} & \left. \frac{d}{dt}\log(1+t) \right|_{t=0} - \left. \frac{d}{dt}\log(1+3t) \right|_{t=0} \\ & = & \frac{1}{1+0} - \frac{3}{1+3 \cdot 0} = \boxed{-2} \end{eqnarray*}

Another way is using

• $(1-y)^{\frac{1}{y}}\stackrel{y \to 0}{\longrightarrow} e^{-1}$

So, \begin{eqnarray*}x^2\log\bigg(\dfrac{x^2+1}{x^2+3}\bigg) & \stackrel{x^2 = \frac{1}{t}}{=} & \frac{\log \frac{1+t}{1+3t}}{t}\\ & = & \log \left(1 - \frac{2t}{1+3t}\right)^{\frac{1}{t}} \\ & = & \log \left( \left(1 - \frac{2t}{1+3t}\right)^{\frac{1+3t}{2t}}\right)^{\frac{2t}{1+3t}\cdot \frac{1}{t}} \\ & \stackrel{t \to 0^+}{\longrightarrow} & \log \left(e^{-1}\right)^2 = \boxed{-2}\\ \end{eqnarray*}

Answer 8

$$x^{2}\ln\left(\frac{x^{2}+1}{x^{2}+3}\right)$$$$=\ln\left(\lim\limits_{x\to+\infty}\left(1+\left(\frac{x^{2}+1}{x^{2}+3}-1\right)\right)^{x^{2}}\right)$$$$=\ln\exp\left(-2\lim\limits_{x\to+\infty}\frac{x^{2}}{x^{2}+3}\right)=-2$$