explain the solution and/or suggest a different one

Question

I have come across the following problem, in my calculus II course, about improper integrals:

problem: Find the following limit, if it exists.

$\displaystyle\lim_{x\to 1} \int\limits_{x}^{x^2} \! \bigg\{ \frac{1}{\log(t)}-\frac{1}{t\log(t)} \bigg\} \, \mathrm{d}t$

$\log(t)$ is the natural logarithm of $t$.

The suggested solution is the above:

solution: Since $\displaystyle \lim_{t\to 1} \bigg\{ \frac{1}{\log(t)}-\frac{1}{t\log(t)} \bigg\}= \cdots= 1$, we get that $\displaystyle\lim_{x\to 1} \int\limits_{x}^{x^2} \! \bigg\{ \frac{1}{\log(t)}-\frac{1}{t\log(t)} \bigg\} \, \mathrm{d}t = 0$

I am having a hard time understanding the solution. What is the thoughts that lead as to calculate the $\displaystyle \lim_{t\to 1} \bigg\{ \frac{1}{\log(t)}-\frac{1}{t\log(t)} \bigg\}$? Should we consider the fact that $\displaystyle \lim_{x\to 1} x =\lim_{x\to 1} x^2 =1$? Is this approach suitable for another problem? Is there any other approach?

Moreover i would like to see more examples, texts on improper integrals with variable bounds, and limits with improper integrals, if there are any.

Thank you in advance.

Answer 1

The limit calculation shows that the integrand is bounded close to $1$. And the integral is now an integral of a bounded function over an interval whose length tends to zero. Thus, the limit is zero. Try go convince yourself that this is true 1) by drawing a graph and 2) by completing the argument with $\epsilon$ and $\delta$ if that is what you are into...

Answer 2

$$\lim_{t\to 1}\dfrac{1}{\ln{t}}-\dfrac{1}{t\ln{t}}=\lim_{t\to 1}\dfrac{t-1}{t\ln{t}}=\lim_{t\to 1}\dfrac{1}{t}\cdot\lim_{t\to 1}\dfrac{t-1}{\ln{t}}=\lim_{t\to 1}\dfrac{t-1}{\ln{t}}$$ $$\lim_{t\to 1}t-1=\lim_{t\to 1}\ln{t}=0$$ Apply L'Hôpital's rule and differentiate both terms separately: $$\lim_{t\to 1}\dfrac{1}{\ln{t}}-\dfrac{1}{t\ln{t}}=\lim_{t\to 1}\dfrac{1}{\frac{1}{t}}=\lim_{t\to 1}t=1$$