Prove $\lim\limits_{x\rightarrow 1} \frac{x^2+3}{x+1}=2$ using the formal definition of the limit.

My question is, I've picked $\delta\lt1$, and I've found that $\delta \lt \min(1,\sqrt{\epsilon})$. Was picking $1$ problematic at all? and is my choice for $\delta$ correct?

Rest of Proof:

$$0\lt|x-1|\lt \delta \Rightarrow \left|\frac{(x-1)^2}{x+1}\right|\lt \epsilon$$

Picking $\delta \lt 1$:

$$|x-1|\lt 1 \Rightarrow -1\lt x-1 \lt 1$$

And we get from that $\frac13 \lt \frac{1}{x+1} \lt 1$ which leads to $\left|\frac{1}{x+1}\right| \lt 1$

Let's go back:

$$\left|\frac{(x-1)^2}{x+1}\right|\lt \left|1(x-1)^2\right|\lt \epsilon$$

Since $(x-1)^2\gt 0$ we can get rid of the absolute value and we get

$$(x-1)^2\lt \epsilon \rightarrow x-1 \lt \sqrt{\epsilon}$$

Also: What is the difference between picking $\delta=1$ and $\delta \lt 1$

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    $\begingroup$ It's hard to say without seeing the rest of your proof. $\endgroup$ – Théophile Jul 28 '16 at 19:16
  • $\begingroup$ @Théophile fixed. Thanks for suggestion. $\endgroup$ – RonaldB Jul 28 '16 at 19:24
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    $\begingroup$ I think your line 3 is wrong. Line 2 implies that $1/3 < (x+1)^{-1} < 1/2$. $\endgroup$ – harvey Jul 28 '16 at 19:31
  • $\begingroup$ @harvey think I fixed it. $\endgroup$ – RonaldB Jul 28 '16 at 19:33
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    $\begingroup$ I think you made a mistake on one of your computations. Oh forget it,now I see what you did. Clever algebra. $\endgroup$ – Mathemagician1234 Jul 28 '16 at 19:34

It seems me correct,the alternative short proof could be:

For any $\epsilon >0$ ,and by choosing $\quad \delta =min\left( 1,\epsilon \right) $ and note that if $\left| x-1 \right| <\delta $ the we will get

$$ \left| \frac { x^{ 2 }+3 }{ x+1 } -2 \right| =\left| \frac { { x }^{ 2 }-2x+1 }{ x+1 } \right| =\left| \frac { { \left( x-1 \right) }^{ 2 } }{ \left( x-1 \right) +2 } \right| <\left| \frac { { \left( x-1 \right) }^{ 2 } }{ \left( x-1 \right) } \right| =\left| x-1 \right| <\epsilon $$

  • $\begingroup$ Thanks! Could you elaborate on why you picked $\delta=*$ and not $\delta \lt*$ Also wow that solution is a beauty $\endgroup$ – RonaldB Jul 28 '16 at 19:37
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    $\begingroup$ you 're welcome.i think $\delta <1$ also posible.may be because it gives you exact answer we usually write $\quad \quad \delta =min\left( 1,\epsilon \right) $ $\endgroup$ – haqnatural Jul 28 '16 at 19:44
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    $\begingroup$ If you pick $\delta = \epsilon$, you can relax the first inequality to $\leq$. $\endgroup$ – harvey Jul 28 '16 at 19:44

Here is a more "gimmicky" kind of solution (Battani's is definitely more elegant and natural): We find a positive constant $C$ such that $\left|\frac{x-1}{x+1}\right|<C\Rightarrow |x-1|\left|\frac{x-1}{x+1}\right|<C|x-1|$, and we can make $C|x-1|<\epsilon$ by taking $|x-1|<\frac{\epsilon}{C}=\delta$. We restrict $x$ to lie in the interval $|x-1|<1$ and note the following: \begin{align} |x-1|<1&\implies 0<x<2\\[1em] &\implies 1<x+1<3\\[1em] &\implies 1>\frac{1}{x+1}>\frac{1}{3}\\[1em] &\implies 3>\frac{x-1}{x+1}\\[1em] &\implies C=3. \end{align} Thus, we should choose $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. To see that this choice of $\delta$ works, consider the following: Given $\epsilon>0$, we let $\delta=\min\left\{1,\frac{\epsilon}{3}\right\}$. If $|x-1|<1$, then $\left|\frac{x-1}{x+1}\right|<3$. Also, $|x-1|<\frac{\epsilon}{3}$. Hence, $$ \left|\frac{x^2+3}{x+1}-2\right|=|x-1|\left|\frac{x-1}{x+1}\right|<\frac{\epsilon}{3}\cdot3=\epsilon, $$ as desired. $\blacksquare$


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