Continuity of $1/x^2$ from the $\varepsilon$ – $\delta$ definition¶

Math I: Calculus · Mock exam, Question 2(b) · 14 points · Lecture 5

Problem¶

Use the $\varepsilon$ – $\delta$ definition to prove that

f\colon\mathbb{R}\setminus\{0\}\to\mathbb{R},\quad f(x)=\frac{1}{x^2},

is continuous at every point of its domain.

What you need¶

Continuity: $f$ is continuous at $x_0$ if for every $\varepsilon>0$ there is a $\delta>0$ such that $|x-x_0|<\delta$ implies $|f(x)-f(x_0)|<\varepsilon\text{.}$
Triangle inequality, in both directions: $|a+b|\leq|a|+|b|$ and $|a|\geq|b|-|a-b|\text{.}$
Absolute value rules: $|ab|=|a|\,|b|$ and $|x^2|=|x|^2\text{.}$

Plan¶

The proof has one idea: write $|f(x)-f(x_0)|$ as $|x-x_0|$ times something, then keep that "something" bounded by staying close to $x_0\text{.}$ Close to $x_0$ means two things here: $x$ must stay away from $0\text{,}$ where $1/x^2$ blows up, and $x$ must not be too large. A first restriction $|x-x_0|<|x_0|/2$ gives both. The final $\delta$ is the smaller of that restriction and the one that makes the product less than $\varepsilon\text{.}$

Step-by-step solution¶

Try each step yourself before you open it.

Step 1 · Fix the point and the tolerance (1 point)

Let $x_0\neq0$ be arbitrary and let $\varepsilon>0\text{.}$ Everything that follows may depend on $x_0$ and $\varepsilon\text{,}$ but not on $x\text{.}$

Writing "let $x_0\neq0$ be arbitrary" at the start is what lets you conclude continuity at every point at the end.

Step 2 · Factor the difference (2 points)

For $x\neq0\text{,}$

\left|\frac{1}{x^2}-\frac{1}{x_0^2}\right| =\frac{|x_0^2-x^2|}{x^2x_0^2} =\frac{|x-x_0|\,|x+x_0|}{x^2x_0^2}.

The factor $|x-x_0|$ is the one we control with $\delta\text{.}$ The other factor, $|x+x_0|/(x^2x_0^2)\text{,}$ must be bounded by a constant that depends only on $x_0\text{.}$

Step 3 · Restrict to a neighbourhood that avoids 0 (2 points)

Require first that

|x-x_0|<\frac{|x_0|}{2}.

Any fixed fraction of $|x_0|$ smaller than $1$ works; $1/2$ keeps the numbers simple. This interval does not contain $0\text{,}$ so every such $x$ lies in the domain.

Step 4 · Bound the denominator from below (2 points)

By the reverse triangle inequality,

|x|\geq|x_0|-|x-x_0|>\frac{|x_0|}{2},

so $x^2>x_0^2/4\text{.}$ A lower bound on the denominator gives an upper bound on the fraction.

Concept: Triangle inequality

Step 5 · Bound the numerator from above (2 points)

Write $x+x_0=(x-x_0)+2x_0$ and use the triangle inequality:

|x+x_0|\leq|x-x_0|+2|x_0| <\frac{|x_0|}{2}+2|x_0|=\frac{5}{2}|x_0|.

Step 6 · Combine the estimates (2 points)

Put Steps 4 and 5 into Step 2:

|f(x)-f(x_0)| \leq\frac{\tfrac{5}{2}|x_0|}{\tfrac{1}{4}x_0^2\cdot x_0^2}\,|x-x_0| =\frac{10}{|x_0|^3}\,|x-x_0|.

This holds for every $x$ with $|x-x_0|<|x_0|/2\text{.}$ The constant $10/|x_0|^3$ depends on $x_0$ only, which is allowed.

Write $\leq$ here, not $<\text{:}$ the bound on the other factor is strict, but $|x-x_0|$ can be $0$ (take $x=x_0$ $x_{0}$ ), and then both sides are $0\text{.}$ The strict inequality comes in Step 7 from $|x-x_0|<\delta\text{.}$

Step 7 · Choose δ and verify (2 points)

Take

\delta=\min\left\{\frac{|x_0|}{2},\,\frac{\varepsilon|x_0|^3}{10}\right\}.

Then $\delta>0\text{.}$ If $|x-x_0|<\delta\text{,}$ the first entry of the minimum makes Step 6 valid, and the second gives

|f(x)-f(x_0)|\leq\frac{10}{|x_0|^3}\,|x-x_0| <\frac{10}{|x_0|^3}\cdot\frac{\varepsilon|x_0|^3}{10}=\varepsilon.

Show the verification explicitly; choosing $\delta$ alone does not earn this mark.

Step 8 · Conclude (1 point)

Since $x_0\neq0$ and $\varepsilon>0$ were arbitrary, $f$ is continuous at every point of $\mathbb{R}\setminus\{0\}\text{.}$

Where marks are lost¶

Choosing $\delta$ that depends on $x\text{.}$ The value of $\delta$ may use $x_0$ and $\varepsilon$ only.
Forgetting the first restriction. Without a bound like $|x-x_0|<|x_0|/2\text{,}$ the factor $1/x^2$ is unbounded near $0$ and Step 6 fails. The minimum in $\delta$ must contain both restrictions.
Bounding $|x|$ from above instead of below. In the denominator you need a lower bound for $|x|\text{.}$
Stopping at "choose $\delta$ ". The last three marks need the check that $|x-x_0|<\delta$ really gives $|f(x)-f(x_0)|<\varepsilon$ and the conclusion for every $x_0\text{.}$

Try next¶

Continuity and differentiability of $x^2\sin(A/x)$ : continuity at a point with a parameter.
Limit of a sequence from the definition: the same "find a bound, then choose $N$ " idea for sequences.

Symbols: delta, epsilon, real numbers

Continuity of 1/x21/x^21/x2 from the ε\varepsilonε–δ\deltaδ definition¶