From the definition of convergence , for every $\epsilon > 0$ there is $N(\epsilon) \in \mathbb{N}$ such that $(\forall) n \geq N(\epsilon)$ , we have : $$ l-\epsilon < \frac{a_{n+1}-a_n}{b_{n+1}-b_n} < l + \epsilon $$ Because $b_n$ is strictly increasing we can multiply the last equation with $b_{n+1}-b_n$ to get : $$ (l-\epsilon)(b_{n+1}-b_n) < a_{n+1}-a_n < (l+\epsilon)(b_{n+1}-b_n) $$ Let $k>N(\epsilon)$ be a natural number . Summing the last relation we get : $$ (l-\epsilon)\sum_{i=N(\epsilon)}^{k}(b_{i+1}-b_i) < \sum_{i=N(\epsilon)}^{k}(a_{n+1}-a_n) < (l+\epsilon)\sum_{i=N(\epsilon)}^{k}(b_{i+1}-b_i) \Rightarrow $$ $$ (l-\epsilon)(b_{k+1}-b_{N(\epsilon)}) < a_{k+1} - a_{N(\epsilon)} < (l+\epsilon)(b_{k+1}-b_{N(\epsilon)})$$ Divide the last relation by $b_{k+1}>0$ to get : $$ (l-\epsilon)(1 - \frac{b_{N(\epsilon)}}{b_{k+1}}) < \frac{a_{k+1}}{b_{k+1}} - \frac{a_{N(\epsilon)}}{b_{k+1}}<(l+\epsilon)(1 - \frac{b_{N(\epsilon)}}{b_{k+1}}) \Leftrightarrow$$ $$ (l-\epsilon)(1 - \frac{b_{N(\epsilon)}}{b_{k+1}}) + \frac{a_{N(\epsilon)}}{b_{k+1}}< \frac{a_{k+1}}{b_{k+1}}<(l+\epsilon)(1 - \frac{b_{N(\epsilon)}}{b_{k+1}}) + \frac{a_{N(\epsilon)}}{b_{k+1}} $$ This means that there is some $K$ such that for $k \geq K$ we have : $$(l-\epsilon)<\frac{a_{k+1}}{b_{k+1}}< (l+\epsilon)$$ (since the other terms who were left out converge to 0)

This obviously means that : $$ \lim_{n \rightarrow \infty} \frac{a_n}{b_n}=l$$ and we are done .