What is this limit (fundamental theorem of calculus)

Question

$lim_{x\to\ 1} \frac{f(x)}{lnx}$

$f(x)= \int_1^u e^{-sin t}dt$

$u=x^2$

I found out that limit will be $\frac{0}{lnx}=0$ is that correct?

Answer 1

By L'Hôpital's rule, one gets $$ \lim_{x \to 1}f(x)=\lim_{x \to 1}\frac{\left(\int_1^{x^2}e^{-\sin(t)}dt\right)'}{(\ln x)'}=\lim_{x \to 1}\frac{2x \cdot e^{-\sin(x^2)}}{\frac1x}=2e^{-\sin(1)}. $$

Answer 2

Let's use L'Hospital rule.

we have

$f(x)=F(x^2)$ with

$$F(x)=\int_1^x e^{-\sin(t)}dt$$

thus

$$f'(x)=2xF'(x^2)=2xe^{-\sin(x^2)}$$

and the limit is

$$\lim_{x\to1}\frac{f(x)}{\ln(x)}=\lim_{x\to1}xf'(x)=2e^{-\sin(1)}$$