Question

using delta - epsilon definition prove that lim(x-->0) (sinx/x) = 1

Answer 1

Answer on Question #73969 – Math – Real Analysis

Question

Using delta - epsilon definition prove that $\lim(x \to 0) (sinx/x) = 1$

Solution

For $x > 0$: $x - \frac{x^3}{3!} < \sin x < x$.

For $x < 0$: $x < \sin x < x - \frac{x^3}{3!}$.

For $x = 0$: $\sin x = x$.

Thus, $|\sin x - x| \leq \frac{x^3}{3!}$ or $\left| \frac{\sin x}{x} - 1 \right| \leq \frac{x^2}{6}$.

Then, whenever $|x - 0| = |x| < \delta$, $\left|\frac{\sin x}{x} - 1\right| \leq \frac{x^2}{6} < \frac{\delta^2}{6}$.

So, for every $\varepsilon > 0$, no matter how small, we can choose $\delta = \sqrt{6\varepsilon}$, such that $|x - 0| < \delta$ implies $\left|\frac{\sin x}{x} - 1\right| < \varepsilon$ we have proved that $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Answer provided by https://www.AssignmentExpert.com