Proving limits with existing results $\lim_{x\to a} \frac{\sin^2 x - \sin^2 a}{x-a} = \sin 2a$ So in my lecture yesterday I learnt how to prove that $\lim_{x\to +\infty} \left(1 + \dfrac{1}{x}\right)^{x} = e$, but I'm lost as to how to apply the results to prove limits.
Any help would be greatly appreciated.
Use the results $\lim_{x\to 0} \left(\dfrac{\sin x}{x}\right) = 1$ and $\lim_{x\to +\infty} \left(1 + \dfrac{1}{x}\right)^{x} = e $ to show that:
$(i) \lim_{x\to a} \dfrac{\sin^2 x - \sin^2 a}{x-a} = \sin 2a;$
$(ii) \lim_{x\to -\infty} \left(1 + \dfrac{1}{x}\right)^{x} = e;$
$(iii) \lim_{x\to 0} (1 + x)^{\frac{1}{x}} = e$
 A: i) Let $y=x-a$. Then, when $x\to a$, $y\to 0$ and we have 
$$
\lim_{y\to 0}\frac{\sin^2(y+a)-\sin^2{a}}{y} = \lim_{y\to 0}\frac{\sin^2 y\cos^2a+2\sin a\cos a\sin y\cos y +\sin^2 a\cos^2 y-\sin^2{a}}{y} = \lim_{y\to 0}\frac{\sin^2 y\cos^2a+2\sin a\cos a\sin y\cos y -\sin^2 a\sin^2 y}{y}=\lim_{y\to 0}\frac{\sin y\cos a\sin(a+y)+\sin a\sin y\cos(a+y)}{y} = \\
\cos a\underbrace{\lim_{y\to 0}\frac{\sin y \sin(a+y)}{y}}_\text{$\lim \sin (a+y) \lim \frac{\sin y}{y}=\sin a$} + \sin a\underbrace{\lim_{y\to 0}\frac{\sin y \cos (a+y)}{y}}_\text{$\lim \cos(a+y)\lim \frac{\sin y}{y}=\cos a$}=2\sin a\cos a=\sin(2a)
$$
In the step where we break the limits, note that given that $\lim_{y\to 0}\sin(a+y)=\sin a$ and $\lim_{y\to 0}\cos(a+y)=\cos a$, then, when multiplying for $\frac{\sin y}{y}$, the limit of the product becomes the product of the limits.
ii) Let $y=-x$. Then, as $x\to -\infty$, $y\to +\infty$ and we have
$$
\lim_{y\to +\infty} (1-\frac{1}{y})^{-y} = \lim_{y\to +\infty}(\frac{y}{y-1})^{y}=\lim_{y\to+\infty}(\frac{y-1}{y-1}+\frac{1}{y-1})^{y} = \lim_{y\to+\infty}(1+\frac{1}{y-1})^{y} = \lim_{z\to \infty}(1+\frac{1}{z})^{z+1} =\underbrace{\lim_{z\to \infty}(1+\frac{1}{z})}_\text{=1}\lim_{z\to \infty}(1+\frac{1}{z})^z = e
$$
In the last step, we took $z=y-1$, but since $y\to \infty$, $z\to \infty$ as well.
iii) You just make the substitution $y=1/x$. Then, when $x\to 0$, either $y\to \infty$ or $y \to -\infty$, but from the previous item and the definition of $e$, in both cases it goes to $e$.
