I was wondering how I could use the limit of a function definition to prove that $\lim_{x \rightarrow 4} f(x) = 1/4$, where $f$ is defined as $$f(x) = \frac{x+1}{x^2+2 \sqrt{x}}.$$ Here's some work so far.
$$\left| \frac{x+1}{x^2+2 \sqrt{x}} - \frac{1}{4} \right| \leq \frac{\left| x+1 \right|}{ \left| x^2 + 2 \sqrt{x}\right|} + \frac{1}{4} = \frac{\left|x-4\right| + 3}{\left|x^2 + 2 \sqrt x \right|} + \frac{1}{4}$$
Suppose we let $\delta_1 = 5$, so that $1 < \left|x\right| < 9$ and $3 < 2 \sqrt{|x|} + |x|^2 < 87$. Can I say that $$ < \frac{\delta + 3}{3} + \frac{1}{4} $$ and for any $\varepsilon$, choose $\delta = \min ( 5, (12 \varepsilon - 15)/4 )$ to satisfy $0 < |x-4| < \delta$ and $|f(x)-1/4| < \varepsilon$? Also, should I mention that $\delta$ should be chosen so that itself and $\varepsilon$ is greater than 0? Thanks in advance.
You have\begin{align}|f(x)-f(4)|&=\left|\frac{x+1}{x^2+2\sqrt x}-\frac14\right|\\&=\frac{\left|4-2\sqrt x+4x-x^2\right|}{4(x^2+2\sqrt x)}\\&=\frac{\left|4-2\sqrt x+4x-16-x^2+16\right|}{4(x^2+2\sqrt x)}\\&\leqslant\frac{2\left|\sqrt x-2\right|+4|x-4|+|x^2-16|}{4(x^2+2\sqrt x)}\\&=\frac{2\frac{|x-4|}{\sqrt x+2}+4|x-4|+|x+4||x-4|}{4(x^2+2\sqrt x)}\\&=\frac{2\frac1{\sqrt x+2}+4+|x+4|}{4(x^2+2\sqrt x)}|x-4|.\end{align}Now, suppose that $|x-4|<3$. Then $x>1$, and therefore$$x^2+2\sqrt x>3\quad\text{and}\quad\sqrt x+2>3.$$On the other hand, $x<7$, and therefore $|x+4|<11$. So,\begin{align}\frac{2\frac1{\sqrt x+2}+4+|x+4|}{4(x^2+2\sqrt x)}&<\frac{2/3+15}{12}\\&<2.\end{align}Therefore, given $\varepsilon>0$, if you take $\delta=\min\left\{3,\frac{\varepsilon}2\right\}$, then, when $|x-4|<\delta$, you have\begin{align}|f(x)-f(4)|&\leqslant\frac{2\frac1{\sqrt x+2}+4+|x+4|}{4(x^2+2\sqrt x)}|x-4|\\&<2\times\frac\varepsilon2\\&=\varepsilon.\end{align}