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I am trying (unsuccessfully) to find the following limit:

$$ \lim_{t \to \infty} \frac{1}{t^2}\int_0^t\ln(2e^x+x^2)dx $$

Some background - supposedly, this exercise is meant to be similar to what we've done in class, where we found the limit: $$\lim \limits_{t \to 0} \frac{1}{t}\int _1^e \ln \left(1+\frac{t}{x}\right)dx$$

We started by simplifying the integral to: $$ \int _1^e \ln \left( x+t \right) dx- \int _1^e \ln \left( x \right) dx$$

and using change of variables and the integral's additivity, arrived at the following: $$ \int _e^{e+t} \ln(y)dy - \int _1^{1+t} \ln(x)dx $$

At this point we've used the mean value theorem, namely that there exist $c_t \in [e,e+t] $ and $d_t \in [1,1+t] $ such that:

$$ \int _e^{e+t} \ln(y)dy - \int _1^{1+t} \ln(x)dx = (e+t-e)\ln(c_t) - (1+t-1)\ln(d_t) = t\left(\ln(c_t) - \ln(d_t)\right) $$

and therefore our limit is:

$$\lim \limits_{t \to 0} \frac{1}{t} \int _1^e \ln \left(1+\frac{t}{x}\right)dx = \lim_{t \to 0} [\ln(c_t) - \ln(d_t)] = \ln(e) - \ln(1) = 1$$

Where we used the squeeze theorem and ln's continuity for the last steps.

I am unsure how to use the mean value theorem when $ t \to \infty $, I've tried using the change of variables $ y = \frac{1}{x} $ in order to turn the upper limit from $t$ to $\frac{1}{t}$ (which approaches zero similarly to what we've done in class), but this also means that the lower limit switches from $0$ to $\infty$ so I didn't see how this method would work.

I would appreciate any and all help/advice.

Daniel Leza
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1 Answers1

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This method does not involve mean value theorem:

$$ \lim_{t \to \infty} \frac{1}{t^2}\int_0^t\ln(2e^x+x^2)dx $$

Note that $2e^x+x^2$ is an increasing function. Therefore it follows that $\ln(2e^x+x^2)$ is an increasing function as well.

This would mean that if the upper limit tends to $\infty$, then the integral $\to \infty$ as well

Now note that the expression is of the form $\frac{\infty}{\infty}$

HINT 1:

Use L'Hopital's rule with Leibniz's rule

HINT 2:

$$ \lim_{t \to \infty} \frac{1}{t^2}\int_0^t\ln(2e^x+x^2)dx = \lim_{t \to \infty} \frac{\ln(2e^t+t^2)}{2t} $$Apply L'Hopital's rule again

Regarding your last question, you could switch the limits using the fact that

$$\int_a^b f(x)dx=-\int_b^a f(x)dx$$

DatBoi
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