So I'm trying to evaluate this limit without using squeeze theorem and doing it by $ \epsilon – \delta$ definition.
Here's my attempt:
$$ \left| \frac{\sin(x)}{x} – 1 \right| \leq \left| \frac{\sin(x)}{x}\right| + |1| \leq \frac{1}{|x|} + 1 $$
Let $\epsilon > 0 $ and $ \delta = \frac{1}{\epsilon}$
$$ |x| < \delta \Rightarrow \frac{1}{|x|} + 1 < \epsilon + 1$$
And I'm stuck here. I also tried $\delta = \frac{1}{\epsilon – 1} $ but that won't work since I'll only proof for $\epsilon > 1$.
Can someone help me? Thanks.
Best Answer
Hint:
Since $\cos\theta<\frac{\sin\theta}{\theta}<1$ $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|<1-\cos\theta$$ and $1-\cos\theta=2\sin^2\frac{\theta}{2}\leqslant\frac{\theta^2}{2}$ hence $$\bigg|\frac{\sin\theta}{\theta}-1\bigg|\leqslant\frac{\theta^2}{2}$$