Prove that $\lim_{k\to\infty}\frac{\sqrt k}{k+7} = 0$ Please I need help on how to prove by definition of a limit. I have been trying since last night but with no luck.
Prove by definition that 
$$\lim_{k\to\infty}\frac{\sqrt k}{k+7} = 0$$
 A: Start by writing down exactly what it is that you need to prove: for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that 
$$\left|\frac{\sqrt k}{k+7}\right|<\epsilon\quad\text{whenever}\quad k\ge n_\epsilon\;.$$ Clearly in this case the absolute value isn’t doing anything, so we really just want to find $n_\epsilon$ such that $$\frac{\sqrt k}{k+7}<\epsilon\quad\text{whenever}\quad k\ge n_\epsilon\;.\tag{1}$$
The function $\dfrac{\sqrt k}{k+7}$ is just barely simple enough that you could try working backwards: solve the inequality $\dfrac{\sqrt k}{k+7}<\epsilon$ for $k$ to see just how big $n_\epsilon$ has to be to make $(1)$ true. I’ll show how you might work your way through this one without doing anything fancy, just by following your nose, so to speak.
$$\begin{align*}
\frac{\sqrt k}{k+7}<\epsilon\quad&\text{iff}\quad\sqrt k<(k+7)\epsilon\\
&\text{iff}\quad\sqrt k<\epsilon k+7\epsilon\\
&\text{iff}\quad\epsilon k-\sqrt k+7\epsilon>0\;.
\end{align*}$$
Let $x=\sqrt k$, and consider the quadratic inequality $\epsilon x^2-x+7\epsilon>0$. It’s convenient to divide through by $\epsilon$ to get $x^2-\frac1{\epsilon}x+7>0$ and then complete the square:
$$\left(x-\frac1{2\epsilon}\right)^2-\frac1{4\epsilon^2}+7>0\;,$$
which can be written more usefully as $$\left(x-\frac1{2\epsilon}\right)^2+7>\frac1{4\epsilon^2}\;.$$ In terms of $k$ that’s $$\left(\sqrt k-\frac1{2\epsilon}\right)^2+7>\frac1{4\epsilon^2}\;.\tag{2}$$
Now $(2)$ will certainly be true if $$\left(\sqrt k-\frac1{2\epsilon}\right)^2>\frac1{4\epsilon^2}\;.\tag{3}$$ Assuming that $\sqrt k\ge\dfrac1{2\epsilon}$, we can take square roots: $(3)$ is true if $\sqrt k\ge\dfrac1{2\epsilon}$ and $\sqrt k-\dfrac1{2\epsilon}>\dfrac1{2\epsilon}$, i.e., if $\sqrt k\ge\dfrac1{2\epsilon}$ and $\sqrt k>\dfrac1\epsilon$.
Of course if $\sqrt k>\dfrac1\epsilon$, then automaticall $\sqrt k\ge\dfrac1{2\epsilon}$, so we now know that $(3)$ (and hence $(2)$) holds whenever $\sqrt k>\dfrac1\epsilon$ or, equivalently, whenever $k>\dfrac1{\epsilon^2}$.
This says that if we let $$n_\epsilon=\left\lceil\frac1{\epsilon^2}\right\rceil\;,$$ the smallest integer $m$ such that $m\ge\dfrac1{\epsilon^2}$, then $(1)$ will be true.
With just a little more experience you’d very likely realize right away that $$\frac{\sqrt k}{k+7}<\frac{\sqrt k}k=\frac1{\sqrt k}\;,$$ so that all you really need is to find $n_\epsilon$ big enough so that $$\frac1{\sqrt k}<\epsilon\quad\text{whenever}\quad k\ge n_\epsilon\;;$$ this would have led you to the inequality $\sqrt k>\dfrac1\epsilon$ very quickly. With a more complicated problem, however, you might very well have to go through the kinds of manipulations that I used in my first solution.
A: Note that $k+7 > k$. Hence, $$\displaystyle \dfrac{\sqrt{k}}{k+7} < \dfrac{\sqrt{k}}{k} = \dfrac1{\sqrt{k}}$$
Now given $\epsilon > 0$, choose $N(\epsilon) = \left \lceil \dfrac1{\epsilon^2} \right \rceil$. Hence, $\forall k > N(\epsilon)$, we have that $k > \dfrac1{\epsilon^2}$ i.e. $\sqrt{k} > \dfrac1{\epsilon}$. Hence,
$$0 < \displaystyle \dfrac{\sqrt{k}}{k+7} < \dfrac1{\sqrt{k}} < \epsilon$$
Hence, $$\lim_{k \to \infty} \displaystyle \dfrac{\sqrt{k}}{k+7} = 0$$
A: Divide numerator and denominator by $\sqrt{k}$. The expression becomes
$$
\frac1{\sqrt{k}+\frac7{\sqrt{k}}}.
$$
What can you say about its individual terms as $k\to\infty$?
