Delta-epsilon proof as $x\to \infty$

Question

Doing my Calculus homework came across a "problem" regarding the applicability of the solution. Using the delta-epsilon definition I tried to prove these two limits. Here is my work:

$$ \lim_{x\to \infty} \frac{1}{x+2}=0\\ \\\forall\epsilon>0\hspace{2mm} \exists M>0\hspace{1mm}\text{such that} \hspace{1mm} x>M \implies|f(x) - L|<\epsilon$$ $$ |\frac{1}{x+2}-0|<\epsilon\\|\frac{1}{x+2}|<\epsilon\\\frac{1}{x+2}<\epsilon\\x+2>\frac{1}{\epsilon}\\x>\frac{1}{\epsilon}-2=M $$ And the second one is: $$ \lim_{x\to \infty} \frac{1}{x^3+2}=0\\|\frac{1}{x^3+2}-0|<\epsilon\\|\frac{1}{x^3+2}|<\epsilon\\\frac{1}{x^3+2}<\epsilon\\x^3+2>\frac{1}{\epsilon}\\x>\sqrt[3]{\frac{1}{e}-2}=M $$

Now, both the solutions seem right, but what does not make sense to me is the thing that according to the definition for all $\epsilon>0$ there has to be $M>0$ and in both cases, there are values for $\epsilon$ when $M$ will be negative. This happens always when there is addition in the denominator. Is there a "special" way of solving this type of task?

Answer 1

There is nothing to “solve” here. You are supposed to prove that, for each $\varepsilon>0$, there is some $M>0$ such that$$x>M\implies\bigl|f(x)-L\bigr|<\varepsilon.\tag1$$If it turns out that, in some cases, you get a $M\leqslant0$ for which $(1)$ holds, then $(1)$ will still be true if, say $M=1$.

Answer 2

You’re missing the fact that the $M$ which you’ve solved for is the minimum value such that $x>M \implies |f(x)-L|<\epsilon$.

For some choices of $\epsilon$, this minimum value might be negative, which simply means that any positive value of $x$ (and even some negative ones) would put $f(x)$ within $\epsilon$ of $L$. You could even have that there is no such minimum value - for example, if you were looking at the limit of $f(x)=\exp(-x^2)$ and chose $\epsilon =2$, then you’d find that $|f(x)|<\epsilon$ for every value of $x$.

The key question is whether you will be able to find some $M$ which works no matter which $\epsilon$ you pick, and you have shown that to be the case in both of your examples.

Answer 3

Sure, set $M=\max\{1,\frac1\epsilon-2\}$. We have two cases:

Case 1: $M=1$. In this case, we have $M>0$ (as needed), and $M\geq\frac1\epsilon-2$. So,

\begin{align*} \left|\frac{1}{x+2}\right|<\left|\frac1{M+2}\right|\leq\left|\frac1{\frac1\epsilon-2+2}\right|=\epsilon \end{align*}

Case 2: $M=\frac1\epsilon-2$. In this case, $\frac1\epsilon-2>1$, so $M>0$, as needed. So,

\begin{align*} \left|\frac{1}{x+2}\right|<\left|\frac1{M+2}\right|=\left|\frac1{\frac1\epsilon-2+2}\right|=\epsilon \end{align*}

Answer 4

When you learnt to solve equations in high school, you might have done something like this:

$$\begin{array}{rcl} \sqrt{x+1} - x + 1 &=& 0 \\ \sqrt{x+1} &=& x-1 \\ x+1 &=& (x-1)^2 \\ x^2 - 3x &=& 0 \\ x &=& 0 \text{(rej.) or } 3 \end{array}$$

It is important to understand what the underlying logic here is.

When we write one equation following the other, we're actually saying that the current equation implies the next equation.

For example, $\sqrt{x+1} = x-1$ implies $x+1 = (x-1)^2$, because if two numbers are equal, then their squares are equal.

However, $x+1 = (x-1)^2$ does not imply $\sqrt{x+1} = x-1$! This is exactly why the solution $x=0$ is rejected: because $\sqrt{0+1} = 1 \ne -1$ even though $1^2 = (-1)^2$.

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However, when you're solving this type of delta-epsilon questions, it is very important to notice that the logic is reversed, since you started at the right hand side of the implication $$x > M \implies |f(x) - L| < \varepsilon$$

In other words, instead of trying to find out what $|f(x) - L| < \varepsilon$ implies, you're actually trying to find out sufficient conditions for $|f(x) - L| < \varepsilon$ to be true.

This is significant in the step $\left|\frac1{x+2}\right| < \varepsilon \impliedby \frac1{x+2} < \varepsilon$, where the arrow is not reversible.

You might find out that a sufficient condition is $x > -1$, i.e. $x > -1$ implies $|f(x) - L| < \varepsilon$.

This poses no problem, as you can simply set $M = 1000$ in this case: if we know that $|f(x) - L| < \varepsilon$ for any $x > -1$, then surely $|f(x) - L| < \varepsilon$ still holds for any $x > 1000$.