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.
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.