$$\lim_{x \to 0} \frac{1-\cos(3x)}{x^2}$$
$$\lim_{x \to 0} \frac{1-\cos(3x)}{1-\cos(2x)}$$
$$\lim_{x \to 0} \frac{\cos(x)-\cos(3x)}{\cos(x)-\cos(2x)}$$
Hi my highschool teacher didn't explain how to solve these limits. I know that if you have something like this you solve it this way: $$ \frac{1-\cos(3x)}{1+\cos(3x)} \cdot \frac{1+\cos(3x)}{x^2} $$ Is this a method of solving the first limit without taking the derivative or using L'Hospital? Do I apply the same rule for the second and third limit? I've seen this technique in a textbook but I don't understand why it was used like that. Can someone explain why you multiply the first equation with $ 1+\cos(3x) $?
The idea is that since the limit of $1+\cos x$ is "nice" (not zero), you can introduce it into the denominator without causing division by zero in the limit (and of course you have to also introduce it into the numerator, so you aren't really changing the original expression).
And it has the particularly nice property that, when combined with an already-existing $1-\cos x$, it becomes $1-\cos^2 x$, which is the same thing as $\sin^2 x$. That's a much better thing to have in a fraction with $x^2$, for instance.
Same thing is true for $1-\cos kx$: you would introduce $1+\cos kx$ in numerator and denominator so they combine to give $1-\cos^2kx = \sin^2 kx$ on one side and just the innocuous $1+\cos kx$ on the other.
For example: $$\frac{1-\cos kx}{\textrm{something}}= \frac{1-\cos kx}{\textrm{something}}\cdot\frac{1+\cos kx}{1+\cos kx}$$ $$=\frac{1-\cos^2 kx}{(1+\cos kx)\cdot{\textrm{something}}}$$ $$=\frac{\sin^2 kx}{\underbrace{(1+\cos kx)}_{\textrm{nice, leave it}}\cdot\underbrace{\textrm{something}}_{\textrm{work with numerator}}}$$