# Thread: Limit epsilon -delta

1. ## Limit epsilon -delta

Hello,
1)lim x goes 1 sgrt(3-2sin(pix/2)=1 prove that epsilon delta.
2)lim x goes pi/4 (sinx+cosx)=sgrt(2) prove that epsilon delta
3)lim x goes3 ((x^2)-9)[[3x]]=0 prove that epsilon delta.
thank you..

2. ## Re: Limit epsilon -delta

"Prove that epsilon delta"?? What does that mean? In any case, please show what you have tried. Do you know the "epsilon-delta" definition of "limit"?

3. ## Re: Limit epsilon -delta

Originally Posted by vaveyla
Hello,
1)lim x goes 1 sgrt(3-2sin(pix/2)=1 prove that epsilon delta.
2)lim x goes pi/4 (sinx+cosx)=sgrt(2) prove that epsilon delta
3)lim x goes3 ((x^2)-9)[[3x]]=0 prove that epsilon delta.
thank you..
Assuming you mean, for example, to prove that \displaystyle \displaystyle \begin{align*} \lim_{x \to \frac{\pi}{4}} \left( \sin{x} + \cos{x} \right) = \sqrt{2} \end{align*} using an \displaystyle \displaystyle \begin{align*} \epsilon - \delta \end{align*} argument...

You should know that if \displaystyle \displaystyle \begin{align*} |x - c| < \delta \implies |f(x) - L| < \epsilon \end{align*} then \displaystyle \displaystyle \begin{align*} \lim_{x \to c}f(x) = L \end{align*}.

So in this case, you need to show \displaystyle \displaystyle \begin{align*} \left| x - \frac{\pi}{4} \right| < \delta \implies \left| \sin{x} + \cos{x} - \sqrt{2} \right| < \epsilon \end{align*}, or if you like, to show

\displaystyle \displaystyle \begin{align*} \left| x -\frac{\pi}{4} \right| < \delta \implies \left| \sin{x} + \cos{x} - \left( \sin{\frac{\pi}{4}} + \cos{\frac{\pi}{4}} \right) \right| < \epsilon \end{align*}

This particular proof requires some knowledge of the Mean Value Theorem, in other words, that \displaystyle \displaystyle \begin{align*} \frac{f(B) - f(A)}{B - A} = f'(C) \end{align*} for some \displaystyle \displaystyle \begin{align*} C \in \left[ A, B \right] \end{align*}

Here our \displaystyle \displaystyle \begin{align*} f(x) = \sin{x} + \cos{x} \end{align*} and \displaystyle \displaystyle \begin{align*} f'(x) = \cos{x} - \sin{x} \end{align*}, so by the Mean Value Theorem

\displaystyle \displaystyle \begin{align*} \frac{f(x) - f\left( \frac{\pi}{4} \right)}{x - \frac{\pi}{4}} &= \cos{C} - \sin{C} \textrm{ for some }C \in \left[x, \frac{\pi}{4}\right] \\ \left| \frac{f(x) - f\left(\frac{\pi}{4}\right)}{x - \frac{\pi}{4}} \right| &= \left| \cos{C} - \sin{C} \right| \\ \frac{\left| \sin{x} + \cos{x} - \sqrt{2} \right|}{\left| x - \frac{\pi}{4} \right|} &= \left| \cos{C} - \sin{C} \right| \\ \frac{\left| \sin{x} + \cos{x} - \sqrt{2} \right| }{\left| x - \frac{\pi}{4} \right| } &\leq \sqrt{2} \textrm{ since }\left| \cos{C} - \sin{C} \right| \leq \sqrt{2} \\ \left| \sin{x} + \cos{x} - \sqrt{2} \right| &\leq \sqrt{2}\left| x - \frac{\pi}{4} \right| \end{align*}

Therefore we should choose \displaystyle \displaystyle \begin{align*} \delta = \frac{\epsilon}{\sqrt{2}} \end{align*}.