Limit of an integral (using Leibniz)

When I try this limit I end up with -1/2 but the answer should be -1, anyone any idea?

$$\lim _{a\to 0}\left(\frac{1}{a²}\int _a^{a²}\frac{\sin\left(ax\right)}{x}\,dx\right)$$

Let $$F(x)=\int_0^x \frac{\sin t}{t}\,dt$$ $$F(0)=0,\,F'(0)=1$$ then, \begin{align} &\lim _{a\to 0}\,\frac{1}{a²}\int _a^{a²}\frac{\sin\left(ax\right)}{x}\,dx\\ (ax \rightarrow t)\,=&\lim_{a\to 0}\,\frac{1}{a^2}\int _{a^2}^{a^3}\frac{\sin\left(t\right)}{t}\,dt\tag1\\ =&\lim _{a\to 0}\left(\frac{F(a^3)}{a^2}-\frac{F(a^2)}{a^2}\right)\\ =&\lim _{a\to 0}\left(a\frac{F(a^3)-F(0)}{a^3}-\frac{F(a^2)-F(0)}{a^2}\right)\\ =&\lim _{a\to 0}\left(aF'(0)-F'(0)\right)\\ =&-1 \end{align}

If using l'Hôpital, note that $$\dfrac{d}{da}\left(\int_{h(a)}^{f(a)}g(x)\,dx\right)=f'(a)g(f(a))-h'(a)g(h(a))\tag2$$ because, if $$G(x)=\int g(x)dx$$ then, $$\int_{h(a)}^{f(a)}g(x)\,dx=G(f(a))-G(h(a))$$ so, going from (1) using (2), \begin{align} \lim_{a\to 0}\,\frac{\int _{a^2}^{a^3}\frac{\sin\left(t\right)}{t}\,dt}{a^2} &=\lim_{a\to 0}\,\frac{\left(\int_{a^2}^{a^3}\frac{\sin t}{t}\,dt\right)'}{2a}\\ &=\lim_{a\to 0}\,\frac{3a^2\frac{\sin {a^3}}{a^3}-2a\frac{\sin {a^2}}{a^2}}{2a}\\ &=\lim_{a\to 0}\,\left(\frac32a\frac{\sin {a^3}}{a^3}-\frac{\sin {a^2}}{a^2}\right)\\ &=-1 \end{align}