Calculating limits of Trig Functions

Question

I'm currently practicing how to calculate the limit of trigonometric functions (without using L'Hospitaler though), and am stuck on problems with additions/subtractions in the denominator. For example:

$$\lim_{x\to0} \frac{1 − \cos 2x}{x^3 + 2x}$$

Where I multiply by:$$\frac{1 + \cos2x }{1 + \cos2x }$$To get:$$ \frac{1 - \cos^2 (2x)}{ 2\cos^2x (x^3 + 2x)} = \frac{\sin^2(2x)}{2\cos^2x (x^3 + 2x)} $$

And get stuck here. . .

Or:

$$\lim_{x\to\pi}\frac{ \sin x}{ x^2 - \pi^2}$$

Any hints, tips, and tricks welcome! Thanks :)

Answer 1

Hint $$\lim_{x\to\pi}\frac{ \sin x}{ x^2-{\pi}^2}\\ =\lim_{x\to\pi}\frac{ -\sin (x-\pi)}{ (x - \pi)(x + \pi)}$$

Answer 2

You can use asymptotic relations to solve this limit. Namely, we know that: $$1-\cos(2x)\sim 4x^2\,\,\, x\to 0$$ So, we have: $$\lim_{x\to0} \frac{1 − \cos 2x}{x^3 + 2x}\sim_\lim{x\to 0}\frac{4x^2}{x^3+2x}=\lim_{x\to}\frac{4x^2}{2x}=0$$ Because, when $x\to 0$, we can neglet $x^3$ in neglect to $2x$.

For te second limit, you can write: $$\lim_{x\to\pi}\frac{ \sin x}{ x^2 - \pi^2}=\lim_{x\to \pi}\frac{1}{x+\pi}\cdot \frac{\sin(x)}{x-\pi}$$ Let $t=x-\pi$. When $x\to \pi$, $t \to 0$, so: $$\lim_{x\to\pi}\frac{ \sin x}{ x^2 - \pi^2}=\lim_{x\to \pi}\frac{1}{x+\pi}\cdot \frac{\sin(x)}{x-\pi}=\frac{1}{2\pi}\cdot\lim_{t\to 0}\frac{\sin(t+\pi)}{t}=\frac{1}{2\pi}\cdot\lim_{t\to 0}\frac{\sin(t)}{t}=\frac{1}{2\pi}$$