# Verify the limit using the formal definition

Verify

$$\lim_{x\rightarrow\infty} \frac{(5+x)}{x^2}=0$$

using the formal definition of a limit. I haven't had much trouble with other formal limit questions, but in this case, I can't figure out how to write $$x$$ in terms of $$\epsilon$$. How do you approach this question?

• Would it help to separate it into $\frac{5}{x^2} + \frac{1}{x}$? – Matthew Leingang Mar 4 at 0:36
• Note that limits to infinity are defined a tad bit differently than most other limits are. Basically, let $0<\frac{5+x}{x^2}<\epsilon$, and after solving, you'd find that $$\epsilon x^2-x-5>0$$Which happens when $x>\frac{1+\sqrt{1+20\epsilon}}{2\epsilon}$. Does this give you a sufficient hint? – Don Thousand Mar 4 at 0:40

## 2 Answers

The problem you are facing is because you are afraid to simplify. Use intermediate inequalities first to reduce the complexity, then bound the result.

$$x\to\infty$$ so we can assume $$x>5$$ and then $$(x+5)<2x$$

So we get $$0<\dfrac{5+x}{x^2}<\dfrac 2x<\varepsilon\quad$$ for $$x>\delta$$ where $$\delta=\max(5,\frac 2\varepsilon)$$

You can make it simpler by observing that for large $$x$$, we get that $$\frac{5+x}{x^2} < \frac{6x}{x^2} = \frac{6}{x}$$. Now let $$\varepsilon>0$$. We must show that for $$x>N$$ we have $$\vert \frac{5 + x}{x^2} - 0 \vert < \varepsilon$$. Let $$N= \frac{6}{\varepsilon}$$. We then get $$\vert \frac{5 + x}{x^2} - 0 \vert < \frac{6}{x} \leq \frac{6}{\frac{6}{\varepsilon}} = \varepsilon$$.