Why $\lim_{t\to 0} \frac{t^2\sin ^2}{(t+\text{sint})(t-\text{sint})}=\lim_{t\to 0} \frac{ \sin ^2t}{2\left(1-\frac{\text{sint}}{t}\right)}$

Question

\begin{align*}\lim_{t\to 0} \frac{t^2\sin^2 t}{(t+\text{sint})(t-\text{sint})}=\lim_{t\to 0} \frac{ \sin ^2t}{2\left(1-\frac{\text{sint}}{t}\right)}\end{align*}

well I cann't get the right side from the left side... why?

Answer 1

Note that

$$\displaystyle \begin{align} \frac{t^2\sin^2t}{(t+\sin t)(t-\sin t)}&=\frac{t^2\sin^2t}{t^2\left(1+\frac{\sin t}{t}\right)\left(1-\frac{\sin t}{t}\right)} =\frac{\sin^2t}{\left(1+\frac{\sin t}{t}\right)\left(1-\frac{\sin t}{t}\right)} \\ \end{align}$$

then

$$\displaystyle \begin{align} \lim_{t\to0}\frac{t^2\sin^2t}{(t+\sin t)(t-\sin t)}&=\lim_{t\to0}\frac{\sin^2t}{\left(1+\frac{\sin t}{t}\right)\left(1-\frac{\sin t}{t}\right)} \\ &=\left(\lim_{t\to0} \frac{1}{1+\frac{\sin t}{t}} \right) \left(\lim_{t\to0} \frac{\sin^2t}{1-\frac{\sin t}{t}} \right)\\ &= \frac{1}{2}\lim_{t\to0} \frac{\sin^2t}{1-\frac{\sin t}{t}}\\&=\lim_{t\to0} \frac{\sin^2t}{2\left(1-\frac{\sin t}{t}\right)} \end{align}$$

because $\displaystyle \lim_{t\to0}{\frac{\sin t}{t}}=1$.

Answer 2

Series: We start more or less where you got to. It is convenient, but not necessary, to multiply and divide by $t^2$. So we want $$\lim_{t\to 0}\frac{1}{1+\frac{\sin t}{t}}\frac{\sin^2t}{t^2}\frac{t^2}{1-\frac{\sin t}{t}}.$$ The first two terms are harmless, combined they have limit $\frac{1}{2}$, so we concentrate on the last part.

Note that $$\sin t=t-\frac{t^3}{3!}+\frac{t^5}{5!}-\cdots.$$ Divide term by term by $t$, and subtract the result from $1$. We get $$\frac{t^2}{3!}-\frac{t^5}{5!}+\cdots.$$ Cancel with the $t^2$ on top. We get $$\frac{1}{\frac{1}{3!}-\frac{t^2}{5}+\cdots}.$$ As $t\to 0$, the terms that involve $t$ die, so the limit is $3!$.

Finally, remember about the $\frac{1}{2}$ that we kind of forgot about: our limit is $3$.

L'Hospital: As I mentioned in a comment, there is the term that goes to $\dfrac{1}{2}$. For the rest we are left with $\dfrac{t\sin^2 t}{t-\sin t}$.

Use L'Hospital's Rule. We could blindly apply it $3$ times. Tedious but doable. Or we can think a little each time.

After the first application, we want the limit of $\dfrac{2t\sin t\cos t+\sin^2 t}{1-\cos t}$.

Multiply top and bottom by $1+\cos t$, and replace $1-\cos^2 t$ by $\sin^2 t$.

You will find the rest easy. All that will be used is the limit of $\dfrac{\sin t}{t}$.

Remark: I just noticed that you had trouble getting the right side from the left. We just divide top and bottom by $t^2$. When we divide the top by $t^2$, we get $\sin^2 t$.

When we divide $(t+\sin t )(t-\sin t)$ by $t^2$, do it by dividing each part by $t$. We get $(1+\frac{\sin t}{t})(1-\frac{\sin t}{t})$.