Use the definition of infinite limit to prove $\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$

Prove $$\lim_{x \to 1+} \frac{x}{x^2-1}=\infty$$

And I was given the solution like this: but I could not understand how it removes the complicated terms.

Let $\delta=\min(0.5,\frac{1}{5M})$.

$$\frac{x}{x^2-1}=\frac{x}{(x+1)(x-1)} \geq\frac{0.5}{\left(\frac{1}{5M}\right)(1.5+1)}=5M\times0.2=M$$

I understand the definition of $M-\delta$, but what I don't understand is what is the solution doing, I mean the process of estimation to get rid of the complicated terms. Anyone can enlighten me? thanks!

• meta.matheducators.stackexchange.com/questions/93/… – Timbuc Nov 29 '14 at 15:04
• let x be an arbitrary real number such that 1<x<1+δ – UnusualSkill Nov 29 '14 at 15:04
• The point is to prove that $\dfrac{x}{x^2-1}$ will be as large as you want (greater than any large $M$) provided that $x-1$ is positive and small enough (ie verifies $\;0<x-1<\delta\,$). $\delta$ is chosen in function of $M$ for this purpose (subtract $1$ and add $1$ in the inequalities and deduce the last line). – Raymond Manzoni Nov 29 '14 at 15:44
• @user3437854: You thus obtain that $x> 1,\;\dfrac 1{x+1}> \dfrac 1{1+1.5},\;\dfrac 1{x-1}> 0.5$. If you suppose further that $x-1<\delta$ you'll get that $\dfrac 1{x-1}> \dfrac 1{\delta}$ (the fraction should be written $\dfrac{0.5}{(1.5+1)\dfrac{1}{5M}}\,$ or more logically I think $\dfrac{1}{(1.5+1)\dfrac{1}{2.5\,M}}\,$ but $1>0.5\,$ so that the first one is correct if not optimal...). (the $0.5$ superior limit is a 'security' for small values of $M$ but is linked to the $5$ coefficient or $2.5$ in front of $M$). – Raymond Manzoni Nov 29 '14 at 16:30
• I think yes ($\frac 1{5M}$ is correct too but unneeded I think here). If you replace the value $0.5$ but another value in $(0,1)$ you'll get another constant in front of $M$. – Raymond Manzoni Nov 29 '14 at 16:48

When $M$ is large, it is enough to take $\delta=\frac{1}{2M}$. This is because if we have $\delta=\frac{1}{2M}$ then the numerator is larger than $1$ while the denominator is less than $\left ( 1 + \frac{1}{2M} \right )\frac{1}{2M}$. Hence the quotient is larger than $\frac{2M}{1+\frac{1}{2M}}$. To make this last expression larger than $M$, we need $\frac{1}{2M} \leq 1$, or in other words $M \geq \frac{1}{2}$.

The problem is that we're in control of $\delta$, not $M$, so we can't require $M \geq \frac{1}{2}$. A workaround is to take $\delta = \min \left \{ \frac{1}{2M},1 \right \}$. Then if $M<\frac{1}{2}$ then the numerator is still larger than $1$ and the denominator is still less than 2, so the quotient is still larger than $\frac{1}{2}>M$.

In general, for any $A,\ B\neq0,\ C\neq0,$ $$\frac{A}{\left(\frac 1B\right)C} = \frac{1}{\left(\frac 1B\right)} \cdot \frac AC = B \cdot \frac{A}{C}.$$ Therefore $$\frac{0.5}{\left(\frac{1}{5M}\right)(1.5+1)} = 5M \cdot \frac{0.5}{1.5+1} = 5M \cdot 0.2.$$

(Assuming the left-hand end of the formula above contains the "complicated terms" that we had to "get rid of".)

Given an $M>0$ somebody found by reverse engineering the proposed $\delta:=\min\left\{0.5,{1\over 5 M}\right\}$ would do the job. Now insert after this definition of $\delta$ the phrase "If $1<x<1+\delta$"; saying that you are only considering points $x$ with $|x-1|<\delta$ and $x>1$. Then your text could read as follows:

If $1<x<1+\delta$ then $0<x-1<{1\over 5M}$, and $x+1<1.5$. It follows that $${x\over x^2-1}={x\over(x-1)(x+1)}>{1\over{1\over 5M}\cdot1.5}> M\ .\tag{1}$$ This means that the proposed $\delta$ does exactly what it was set up to do.

Note that, given $M$, such a $\delta>0$ is not determined uniquely, even though people write $\delta(M)$ sometimes. It is experience with such questions that leads us in finding an admissible $\delta$. Here are some hints: The factor $x-1$ in the denominator of $(1)$ indicates that ${1\over\delta}$ should have the order of magnitude at least $M$. Since the other factors amount to ${x\over1+x}<1$ some extra measures have to be taken to guarantee $(1)$. Nobody is claiming that the proposed $\delta$ is the "optimal" or "simplest" $\delta=\delta(M)$ doing the job.

• What I don understand is how to find the proposed delta?can you help? thx! – UnusualSkill Nov 29 '14 at 16:00
• is there any wrong with the solution? because When i restrict δ=0.5>>>0<x-1<0.5>>>>1<x<1.5 >>>>2<x+1<2.5 >>> 1/2.5<1/(x+1)<1/2 why the numerator is 0.5? – UnusualSkill Nov 29 '14 at 16:06

We can first use partial fraction decomposition to $\frac{x}{x^2-1},$ which leads to \begin{gather*} \frac{x}{x^2-1}=\frac{1}{2(x+1)}+\frac{1}{2(x-1)}, \end{gather*} which implies that \begin{gather*}\tag{1} \frac{x}{x^2-1}>\frac{1}{2(x-1)}, \qquad \text{provided } x>-1. \end{gather*}

Let $M>0.$ Set $$\delta=\frac{1}{2M}.$$ For every $x>1,$ we have \begin{align*} 0<x-1<\delta\implies \frac{x}{x^2-1}&>\frac{1}{2(x-1)}\qquad\qquad \text{by }(1)\\ &>\frac{1}{2\delta}=\frac{1}{2\frac{1}{2M}}=M, \end{align*} which, by definition of limit, completes the proof.