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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 $$\lim_{k\to\infty}\frac{\sqrt k}{k+7} = 0$$

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Well, what have you been trying? – Pedro Tamaroff Oct 20 '12 at 19:01
@BabakSorouh I TeXed it, so maybe the TeXation wasn't faithful. – Pedro Tamaroff Oct 20 '12 at 19:03
@Peter: Since she was careful to use brackets in the denominator, I think that maureen probably meant to have the square root in the numerator; however, I hope that she lets us know for sure. (No TeXation without representation?) – Brian M. Scott Oct 20 '12 at 19:05
@BrianM.Scott: And wasn't the brackets used for $floor(...)$? – Babak S. Oct 20 '12 at 19:08
@BrianM.Scott Let's just wait and see. I hope the TeX headquarters are not in Boston. – Pedro Tamaroff Oct 20 '12 at 19:09

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:


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.

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Yea the sqrt is only for the numerator. I ll follow the step u ve stated. Lim – Nero Oct 20 '12 at 19:26
Oops sori guys, there is an error in the question. It's sqrt K/[k+7]. I just noticed it now. So sori please. But do I follow the same procedure? – Nero Oct 20 '12 at 19:30
@maureen: You’ll have to work a bit harder. I’ve edited the question; now I’ll revise my answer. – Brian M. Scott Oct 20 '12 at 19:36
Hello Brian, can I do this?: $\frac{\sqrt k}{k+7}<\epsilon\quad\text{whenever}\quad k\ge n_\epsilon$ so $\frac{k+7}{\sqrt{k}} > \frac{1}{\epsilon}$ but $k+7>\frac{k+7}{\sqrt{k}} > \frac{1}{\epsilon}$ therefore $k+7 > \frac{1}{\epsilon}\implies k>\frac{1}{\epsilon} + 7$ – Marter Js May 11 '14 at 21:38

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, $$\lim_{k \to \infty} \displaystyle \dfrac{\sqrt{k}}{k+7} = 0$$

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Thanx for the answer given. It indeed helped me. Now I ll study it. – Nero Oct 20 '12 at 19:58

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 $k\to\infty$?

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