finding the value of this limit (upper bound integral) $f(x)$ is continuous and $\lim_{x \to 0} \frac{f(x)}{x} = 1$,$$\lim_{x \to 0} \frac{x\int_0^x{f(x-t)dt}}{\int_0^x tf(x-t)dt}$$
I know how to do this when $f(0) \neq 0$, but I don't know where it go wrong when $f(0)=0$
 A: Given that $\lim_{x\to0} \frac{f(x)}{x} = 1$,
$$f(0)=0$$
And, by L'Hospitales rule,
$$f'(0) = 1$$
Now, apply L'Hopitale's rule repeatedly to get rid of the integral sign (using Leibnitz rule):
$$\lim_{x\to0} \frac{x\int_{0}^{x}f(x-t)dt}{\int_{0}^{x}tf(x-t)dt}$$
$$= \lim_{x\to0} \frac{x\int_{0}^{x}f(t)dt}{\int_{0}^{x}(x-t)f(t)dt}$$
$$= \lim_{x\to0} \frac{\int_{0}^{x}f(t)dt+xf(x)}{\int_{0}^{x}f(t)dt+xf(x)-xf(x)}$$
$$= \lim_{x\to0} \frac{\int_{0}^{x}f(t)dt+xf(x)}{\int_{0}^{x}f(t)dt}$$
$$= \lim_{x\to0} (1+\frac{xf(x)}{\int_{0}^{x}f(t)dt})$$
$$= 1+\lim_{x\to0} \frac{xf'(x)+f(x)}{f(x)}$$
$$= 2+\lim_{x\to0} \frac{f'(x)}{(\frac{f(x)}{x})}$$
$$= 2+\frac{1}{\lim \frac{f(x)}{x}}$$
$$= 2+\frac{1}{1}$$
$$=3$$
A: the condition is $f(0) \neq 0$ or $\lim_{x\to0}\frac{f(x)}{x}=1$, not both
here is the right solution when $f(0) \neq 0$
$$\lim_{x\to0} \frac{x\int_0^x{f(x-t)dt}}{\int_0^xtf(x-t)dt}=\\\lim_{x\to0}\frac{x\int_0^x{f(u)du}}{x\int_0^xf(u)du-\int_0^xuf(u)du}=\\
\lim_{x\to0}\frac{\int_0^xf(u)du+xf(x)}{\int_0^xf(u)du}=\\
\lim_{x\to0}=\frac{\frac{\int_0^xf(u)du}{x}+f(x)}{\frac{\int_0^xf(u)du}{x}}$$ since $$
\lim\frac{\int_0^xf(u)du}{x}=f(0)$$
so it's 2$$\\$$
I just don't know where it got wrong when condition change to $\lim_{x\to0}\frac{f(x)}{x}=1$,I guess it's because $f(0)=0$ in the second condition? btw the answer is 3 for second condition
