## improper integrals: use of the L & # 39; Hopital rule in a limit that does not exist

I was trying to solve an improper integral and I had to evaluate the following expression with the upper limit as infinite (the limit is 0) and the lower limit as 0 (the limit I am asking here)

Here is the limit:
$$lim_ {x a 0} (x ln (e ^ {x} -1) -x ^ {2})$$

I introduced the expression in an online limit solver.

The L & # 39; Hopital rule can be used when the limit is in an undetermined form, and when there is the limit in the numerator and the limit in the denominator. The limit of $$frac {1} {x}$$
It does not exist when x approaches 0. (And I think the limit of the logarithm in the numerator does not exist when it approaches 0, unless only from the right? Someone confirm this please)

Is it safe to assume that the boundary solver assumes that x approaches 0 from the right?
Does this mean that the limit only exists when x is approached from the right?

And if this limit does not exist unless it is from the right, is it safe to change the previous limit so that it approaches from the right to solve this integral?

## calculation – improper integral use comparison theorem

I have to determine using the comparison theorem if the following integral rather converges or diverges:

$$int_0 ^ infty frac { ln {x}} {x ^ 2} dx$$

It is evident that this tintegral is improper because the upper limit goes to infinity and is asymptotic in $$x = 0$$. Then I split it into two parts:

$$int_0 ^ infty frac { ln {x}} {x ^ 2} dx = int_0 ^ 1 frac { ln {x}} {x ^ 2} dx + int_1 ^ infty frac { ln {x}} {x ^ 2} dx$$

Of course, the first part has a negative sign and the second, a positive sign.

So, is it possible to determine with the comparison theorem the nature of the integral? In that case, what are the functions to compare?

Also, is it possible that the negative area can be canceled by the positive area?

## calculation and analysis: how to prove that an improper integral converges or diverges?

Verify the improper integral. $$int_ {1} ^ { infty} x cdot left | sin (x ^ 4) cdot without x right | dx$$ converge or diverge, I code

``````To integrate[x*Abs[Sin[x^4]*Sin[x]]{x, 1, infinity}
``````

then feedback is the expression of $$int_ {1} ^ { infty} x cdot left | sin (x ^ 4) cdot without x right | dx$$.It means the improper integral $$int_ {1} ^ { infty} x cdot left | sin (x ^ 4) cdot without x right | dx$$ converge?

## Is there a \$ n in mathbb {N} \$ such that the Riemann integral improper \$ int_0 ^ { infty} | sin (x ^ {n}) | \$ converge?

I know that $$int_0 ^ { infty} sin (x ^ { alpha})$$ converge for $$| α | geq 1$$ that can be attributed to the cancellation of areas and $$int_0 ^ { infty} | sin (x ^ { alpha}) |$$ diverges for $$n = 2$$ Since there is no cancellation of areas. It seems unlikely, but maybe there is a $$n$$ So that the area of ​​oscillations can be made so small that it can be approximated by a geometric series?