Let $$f(t)=(c-1)t^3,$$ $$g(t)=\displaystyle\int _{\sqrt{\pi}}^{t}\left(\sin{(u^2)}-cu^2\right)du,$$ $h(t)=\dfrac{f(t)}{g(t)}$ ($c$ is a constant)
Now, let's assume that $c$ satisfies $g(0)=0$. Is it right to say that when we solve $$\lim\limits_{t \rightarrow 0} h(t), \lim\limits_{t \rightarrow 0} \dfrac{f(t)}{g(t)} = \lim\limits_{t \rightarrow 0} \dfrac{f^{\prime}(t)}{g ^{\prime}(t)} = \lim\limits_{t \rightarrow 0} \dfrac{3(c-1)t^2}{\sin{(t^2)} - ct^2} = \dfrac{3(c-1)}{1-c} = -3$$
?
Furthermore, is there a workaround WITHOUT using the L'Hospital's rule? I tried the following approach but does not work.
$$\lim\limits_{t \rightarrow 0} \dfrac{f(t)}{g(t)} = \lim\limits_{t \rightarrow 0} \cfrac{\cfrac{f(t) - f(0) }{t}}{\cfrac{g(t) - g(0)}{t}} = \dfrac{f^{\prime}(0)}{g^{\prime}(0)}=\dfrac{0}{0}$$