0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Would it help to separate it into $\frac{5}{x^2} + \frac{1}{x}$? $\endgroup$ – Matthew Leingang Mar 4 at 0:36
  • $\begingroup$ 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? $\endgroup$ – Don Thousand Mar 4 at 0:40
0
$\begingroup$

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 $\delta=\max(5,\frac 2\varepsilon)$

$\endgroup$
0
$\begingroup$

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 $\vert \frac{5 + x}{x^2} - 0 \vert < \frac{6}{x} \leq \frac{6}{\frac{6}{\varepsilon}} = \varepsilon$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.