Evaluate $\lim_{t \rightarrow 0} \int_{0}^t \frac{1}{f(u)}du$ Let $f(u)$ be a function such that $\lim_{u \rightarrow 0} f(u)=0$ e.g. $f(u)={u}$. How would I evaluate
$$
\lim_{t \rightarrow 0} \int_{0}^t \frac{1}{f(u)}du
$$
Is this always equal to zero?
My initial instinct is to use L'Hospital but not sure how or even if it makes sense. 
Edit: Assume $f$ is continuous around $0$. Use the Lebesgue integral. I think also assume that $\frac{1}{f(u)}$ is integrable (if this is not required then ignore). Hence, the example $f(u)=u$ is not a good one because it is not integrable.

If $\frac{1}{|f|}$ is integrable near $0$ then $\int_{0}^{t}\frac{1}{f}=\int_{0}^{1}\frac{\chi_{[0,t]}}{f}$, where $\chi_{[0,t]}$ is the indicator function of $[0,1]$. We have that $\frac{\chi_{[0,t]}}{f}\to0$ and $\left|\frac{\chi_{[0,t]}}{f}\right|\leq\left|\frac{1}{f}\right|$. We can apply dominated convergence theorem to get $$\lim_{t\to0}\int_{0}^{t}\frac{1}{f}=\lim_{t\to0}\int_{0}^{1}\frac{\chi_{[0,t]}}{f}=\int_{0}^{1}\lim_{t\to0}\frac{\chi_{[0,t]}}{f}=\int_{0}^{1}0=0$$
It remains to study the case in which $\frac{1}{f}$ is integrable but $\frac{1}{|f|}$ is not.

If we assume this is Henstock-Kurzweil integral, then we don't need $\frac{1}{|f|}$ to be integrable to apply dominated convergence. It is enough to know that $0\leq \frac{\chi_{[0,t]}}{f}\leq \frac{1}{f}$, and that $0$ and $\frac{1}{f}$ are integrable, which we have.
So, it remains to study the case when a less general integral definition is used, e.g. Riemann integral.

If it is Riemann integral ($\int_{0}^{t}\frac{1}{f}$ exists as an improper Riemann integral) then it is Henstock integrable and their values coincide.
