I have some difficulties understanding the last step of the screenshot.
The first step is to multiple the conjugate of the numerator. The second step is to use $1 - \cos^2(x)$ to get $\sin^2(x)$.
How and why did the author splits the $\dfrac{\sin^2x}{x(1+\cos x)}$ into a limit with two functions in it.
Can anyone help to explain the intentions and how it is derived?
On
Adding to Siong Thye Goh's answer:
Remember that "the limit of a product is the product of the limits", when each individual factor's limit exists, so that
$$ \lim_{x \to 0} \frac{\sin(x)}{x} \frac{\sin(x)}{1+\cos(x)}$$
$$ = \lim_{x \to 0} \frac{\sin(x)}{x} \lim_{x \to 0} \frac{\sin(x)}{1+\cos(x)}$$
$$ = 1 \times 0 = 0.$$
The key point here is understanding why $\frac{\sin(x)}{x}$ tends to 1, as $x$ tends to 0, and then applying this limit; there are some great geometric proofs of this, if you just do a search here on this site. (Note: I don't recommend that you verify this limit with l'hospital's rule, since it's very much circular reasoning.)
It is known that $\lim_{x \to 0} \frac{\sin x}{x}=1$ and taking that term out make things clearer as now when we substitue $0$ into the denominator, we no longer get $0$ but the numerator still give us $0$.
Since $\lim_{x \to 0} \frac{\sin x}{1+\cos x}=0$, the overall limit is $0$.