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i kinda dont know how to solve, to be more precise, i can't find a proper $\delta$ the following limits using epsilon-delta notation:

  • $\displaystyle\lim_{x\to 4}\sqrt{x}=2$

  • $\displaystyle\lim_{t\to 7}\frac{8}{t-3}=2$

the first limit, i found on the solutions manual that the $\delta = \frac{1}{2}$ on the first equation.

in the first limit we have, by using the epsilon-delta formula for limits:

$|\sqrt{x} - 2| < \epsilon$

Multiply by the function's conjugate to get a function with the same format as |x - 4|:

$\left| \sqrt{x} - 4 \cdot \left( \frac{\sqrt{x} + 2}{\sqrt{x} + 2} \right) \right|$

Solving the equations, the numerator is a difference of squares and the denominator is resolved according to the distribubility rules. As the denominator of the function is 1, so the conjugate of the function assumes the denominator of the function:

$\left| \frac{x - 4}{\sqrt{x} + 2} \right|$

transform into a product so that the factor $(x - 4)$ is highlighted:

$\left| (x - 4) \cdot \left( \frac{1}{\sqrt{x}} + 2 \right) \right|$

Then I don't know what else can be done, something with the second factor from left to right.
from this point, the book states that $|x - 4| \cdot \left|\frac{1}{\sqrt{x} + 2}\right| < \frac{1}{2} \cdot |x - 4|$

the same with the second limit: in the first limit we have, by using the epsilon-delta formula for limits:

$\left| \frac{8}{t - 3} - 2 \right| < \epsilon$ there is a delta such that $|x - 7| <\delta$

taking the mmc of the denominators, modifying the mmc by the two denominators, and dividing the mmc of the denominators by the old denominator of the two parts of the equation and multiplying the result by the numerators to leave it in the same format as |x - 7|:

$|t - 7| \cdot \left( \frac{2}{|t - 3|} \right)$

From now on I don't know how to find the delta of the function. the book states that $|x - 7| \cdot \left(\frac{2}{x-3}\right) < |x - 7| \cdot \frac{2}{3}$

i dont know how to get to the fractional part of the right side of equation. the book says that i need to make these as small as the first factor, but i dont know how to do it.

Leonardo Fonseca
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    Recent question: Is this a valid proof of the continuity of $f(x)=\sqrt{x}$ at $x=2$, using epsilon-delta definition? – peterwhy Mar 12 '24 at 03:48
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    A lot of these $\varepsilon$-$\delta$ proof attempts stall at a point like this, where $|f(x) - L|<\varepsilon$ boils down to $|x - a| \cdot |\text{Big Ugly Thing}| < \varepsilon$, where we are trying to prove $\lim_{x \to a} f(x) = L$, and $\text{Big Ugly Thing}$ involves some expression of $x$. The trick is to make sure $\text{Big Ugly Thing}$ doesn't get large as $|x - a|$ gets small. In the first case, $\frac{1}{\sqrt{x} + 2} \le \frac{1}{2}$, so $|\sqrt{x} - \sqrt{2}| \le |x - 2|/2$. In the second case, $1/|t-3|$ does get very large, just not around $t = 7$, so limiting $\delta$ is handy. – Theo Bendit Mar 12 '24 at 04:52
  • "... I don't know how to find the delta ... ." There is no such thing as "the delta". You need only find a delta, and there are many choices that will work. The particular formula for $\delta$ that you're trying to derive need some care to obtain, but you're very close. I think you lost a factor of $2$ in one of the calculations, though. – David K Mar 12 '24 at 06:30
  • but how do i do to get the fractional part? why do you equals $\frac{1}{\sqrt{x}+2} \lt \frac{1}{2}$ ? it could be many other numbers which gives the same result. why do i need to equal the equation to 1/2? – Leonardo Fonseca Mar 12 '24 at 07:28
  • " it could be many other numbers which gives the same result." Yes, and you can use any one of then. "why do i need to equal the equation to 1/2?" You don't. You just have to get one small enough. That's exactly what @DavidK was commenting about. – fleablood Mar 12 '24 at 07:32
  • Your edit that $\delta = \frac 12$ is much worse then the original $\delta=\min (2\epsilon, 4)$ and no longer makes any sense. – fleablood Mar 12 '24 at 07:33
  • sorry, i'm new to this site. what is enough to this context? you mean, to approach the variable to the number which x is tending? if i wanted to pinpoint the right answer, the same in the solutions manual, what would it be the right way to think over this problem? if i equals other small number than the $\frac{1}{2}$, it would give me other values. – Leonardo Fonseca Mar 12 '24 at 07:47

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Lets walk through the first one.

We claim: $\lim_\limits{x\to 4} \sqrt x = 2$

Then we must show that the following is true:

For any $\epsilon > 0$ there exists $\delta > 0$ such that when $|x-4| < \delta$ then $|\sqrt x - 2| < \epsilon$

As you note $|\sqrt x - 2| = |(\sqrt x - 2)\frac {\sqrt x + 2}{\sqrt x + 2}| = | \frac {x-4}{\sqrt x + 2}|$

And, from the line above we only care about the case when $|x-4|< \delta$

$|\sqrt x - 2| = |\frac {x-4}{\sqrt x + 2}| < |\frac {\delta}{\sqrt x + 2}|$

And since $\sqrt {x} \ge 0, |\frac {\delta}{\sqrt x + 2}| \le \frac {\delta}{2}$

Putting it all together:

If $\delta \le 2\epsilon$

$|\sqrt {x} - 2| < |\frac {\delta}{\sqrt x +2}|\le \frac {\delta}{2} \le \epsilon$

And we have shown what we need to show. For any $\epsilon$ we have found that there exists a corresponding $\delta$ that does what we need it to do.

user317176
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  • the 2 dividing delta is the limit? – Leonardo Fonseca Mar 12 '24 at 09:43
  • We started with the evaluation of the limit as given. That is $\lim_\limits{x\to 4} \sqrt x = 2$ and then set out to prove that the limit is true. The limit is true if, for any $\epsilon,$ we can find a corresponding $\delta$ that will satisfy the definition of the limit. With a little algebra, we find that $\delta = 2\epsilon$ is a satisfactory $\delta.$ – user317176 Mar 12 '24 at 10:03