$$\lim_{x\to2}\left(x^2+2x-7\right)\ = 1$$
For every $\epsilon > 0$, there exists a $\delta >0$ such that
$|x-2| < \delta \implies |(x^2+2x-7) - 1| < \epsilon$.
Very often, you solve these problems by looking at what $\epsilon$ needs to do and then working backwards to what $\delta$ needs to do. In this case
$$|(x^2+2x-7) - 1| = |x^2+2x-8| = |(x+4)(x-2)| = |x+4|\,|x-2|$$
So, we need to make $|x+4|\,|x-2| < \epsilon$. We know we are going to make $|x-2| < \delta$, but what do we do with $|x+4|$?
\begin{align}
|x-2| < \delta
&\implies 2-\delta < x < 2 + \delta \\
&\implies 4-\delta < x + 4 < 4 + \delta
\end{align}
The trick is to limit the size of $\delta$. There is no fixed limit that you need to use. Just pick one. I think $10$ is a nice round number so I am going to say, suppose $0 < \delta < 2$. Then
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0<\delta<2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (-2<-\delta<0)\\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
\end{align}
You should see that we now solve $10\delta < \epsilon$ for $\delta$. We get $\delta < \dfrac{\epsilon}{10}$. But wait! We made an assumption that $\delta < 2$. That's very easy to fix. Our final formula is
$\delta = \min\left\{2, \dfrac{\epsilon}{10} \right\}$.
Then we get our proof by adding one more line to the previous argument.
\begin{align}
|x-2| < \delta \; \text{and} \; (0 < \delta < 2)
&\implies (2-\delta < x < 2 + \delta) \; \text{and} \; (0 < \delta < 2) \\
&\implies (6-\delta < x+4 < 6+\delta) \; \text{and} \; (0<\delta<2) \\
&\implies 4 < x + 4 < 8 \\
&\implies |x+4| < 10 \\
&\implies |x+4||x-2| < 10\delta \\
&\implies |(x^2+2x-7) - 1| < \epsilon
\end{align}
How can I use the $\epsilon$-$\delta$ definition of limit to prove that the following limit doesn't exist?
$$ \lim_{x\to1} \sin(\frac{1}{x-1}) $$
Let $~\displaystyle f(x) = \sin\left(\frac{1}{x-1}\right).$
I guess that different people attack this problem in different ways. My approach is to establish that no matter how small a neighborhood of $\delta > 0$ is taken around $x = 1$, I will always be able to find distinct values $x_1, x_2$ that are both inside this neighborhood, so that (for example), for a fixed $r > 0$, you have that
$$|f(x_1) - f(x_2)| > 2r.$$
Assume that this has been done. Then, set $\epsilon = r$, and consider whether the function can converge to a limit $L$. The problem is that
$$|f(x_1) - L| + |f(x_2) - L| > 2r = 2\epsilon, \tag1$$
by the triangle inequality. Therefore, in (1) above, at least one of the two LHS terms must be greater than $\epsilon$. Further, by presumption, this will hold no matter how small $\delta$ is taken. This implies that it is impossible for any limit $L$ to exist such that the function converges to $L$.
Therefore, the problem has been reduced to establishing that regardless of how small $\delta > 0$ is taken, I can find $x_1, x_2$ as distinct values such that
- $0 < |x_1 - 1| < \delta.$
- $0 < |x_2 - 1| < \delta.$
- $|f(x_1) - f(x_2)| > (1/2)$ (for example).
Assuming that the above is demonstrated, then (for example) I could take $\epsilon = (1/4)$, and then apply the analysis of the previous section.
For any (fixed) $~\delta > 0,~$ choose $~M \in \Bbb{Z^+},$ such that $\displaystyle ~M > \frac{1}{\delta}.$
Then, set
- $~\displaystyle x_1 = \frac{1}{M\pi} + 1 \implies $
$\displaystyle |x_1 - 1| = \frac{1}{M\pi} < \frac{1}{M} < \delta.$
- Similarly, set $~\displaystyle x_2 = \frac{1}{[M+(1/2)]\pi} + 1.$
So, now, $x_1, x_2$ are distinct elements in a neighborhood of $\delta$ around $x=1$.
Then
- $\displaystyle \frac{1}{x_1 - 1} = M\pi \implies f(x_1) = 0.$
- $\displaystyle \frac{1}{x_2 - 1} = [M + (1/2)]\pi \implies f(x_2) = \pm 1.$
Thus, $~|f(x_1) - f(x_2)| > (1/2),~$ as required.
Best Answer
Now since $\lim_{x\to 0} \frac{x}{\sin(x)} =1$ we may pick $\delta(\epsilon) >0$ such that $|\frac{\frac{x}{2}}{\sin(\frac{x}{2})} -1|< \frac{\sqrt{2}}{2}\epsilon$.
Fix $\epsilon >0$. Let $x>0$ with $|x|<\delta(\epsilon)$ and note (as you did) that:
\begin{align*} \left| \frac{x}{\sqrt{1-\cos(x)}}-\frac{2}{\sqrt{2}}\right| = \left|\frac{2}{\sqrt{2}}\frac{\frac{x}{2}}{\sin(\frac{x}{2})}-\frac{2}{\sqrt{2}} \right| = \frac{2}{\sqrt{2}}\left|\frac{\frac{x}{2}}{\sin(\frac{x}{2})} -1 \right|<\frac{2}{\sqrt{2}} \frac{\sqrt{2}}{2}\epsilon = \epsilon, \end{align*} as required. So in essence you did all the hard work.