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Given the following limit to infinity, I am supposed to prove using epsilon-delta definition that the limit = 3 as x approaches infinity. How do I approach this?

$$\lim_{x\to\infty}\dfrac{6x+1}{2x+1} = 3$$

Bernard
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Naja
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  • Do you know the epsilon-delta definition of limits to infinity? – Ken Hung Aug 29 '20 at 08:24
  • I understand that the definition of limits to infinity is as such: For any epsilon > 0, there exists N such that | f(x) - Limit | < epsilon, whenever x> N. However I am unsure of how to use the epsilon/delta definition to start the proof – Naja Aug 29 '20 at 08:32

2 Answers2

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$\big| \frac{6x+1}{2x+1}-3 \big| < \epsilon \iff \big| \frac{-2}{2x+1} \big| < \epsilon \iff \frac{1}{\epsilon}< |x+\frac{1}{2}|$.

Can you take it from here?

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A limit to infinity for $f:\mathbb R\to\mathbb R$ (with value $l$) means that for all $\epsilon>0$ there is a $\delta>0$ such that if $|x|\geq\delta$ then $|f(x)-l|\leq\epsilon$.

Then you have:

Suppose $x>0$,

$$\big|\frac{6x+1}{2x+1}-3\big|=\frac{2}{|2x+1}|\leq\frac{2}{2|x|}\leq\frac{1}{\delta}.$$

Then you choose $\delta=\frac{1}{\epsilon}$ and you are done.

Suppose instead $x<0$,

$$\big|\frac{6x+1}{2x+1}-3\big|=\frac{2}{|2x+1|}=\frac{2}{|1-2|x||}\leq\epsilon,$$

by choosing $\delta=\frac{1}{\epsilon}+\frac{1}{2}$.

Bellem
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  • Thanks! I copied from above and forgot a \delta. I had corrected that few seconds before your comment. – Bellem Aug 29 '20 at 09:36
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    Perhaps, in the case $x<0$, you should write (supponing $|x|>\frac{1}{2}$):$$\left|\frac{6x+1}{2x+1}-3\right|=\frac{2}{|2x+1|}=\frac{2}{|1-2|x||}=\frac{2}{2|x|-1}\leq\frac{2}{2\delta-1}=\epsilon,$$ by choosing $\delta=\frac{1}{\epsilon}+\frac{1}{2}$. – Angelo Aug 29 '20 at 09:41
  • Thanks! Never do many things together while you are in a hurry ;) Cya. – Bellem Aug 29 '20 at 09:53
  • One is with the modulus. Or you meant before my last correction probably. – Bellem Aug 29 '20 at 09:56