## University Calculus: Early Transcendentals (3rd Edition)

$\lim_{t\to0}g(t)/h(t)$ can still exist and does not have to equal $0$.
1) Even though $g(0)=h(0)=0$, $$\lim_{t\to0}\frac{g(t)}{h(t)}$$ can still exist. Take this example: $g(t)=t^3$ and $h(t)=t$ Here both $g(t)$ and $h(t)$ are defined for all values of $t$ and $g(0)=h(0)=0$. However, $$\lim_{t\to0}\frac{g(t)}{h(t)}=\lim_{t\to0}\frac{t^3}{t}=\lim_{t\to0}t^2=0^2=0$$ So $\lim_{t\to0}g(t)/h(t)$ still exists. 2) Again, $\lim_{t\to0}g(t)/h(t)$ also does not have to be $0$. Another example: $g(t)=t^2-t$ and $h(t)=t$ Both $g(t)$ and $h(t)$ are defined for all values of $t$ and $g(0)=h(0)=0$. However, $$\lim_{t\to0}\frac{g(t)}{h(t)}=\lim_{t\to0}\frac{t^2-t}{t}=\lim_{t\to0}(t-1)=0-1=-1$$ So $\lim_{t\to0}g(t)/h(t)$, in this case, does not equal $0$.