The squeeze theorem and special functions

What you will be able to do

• trap a function between two simpler ones and read off its limit;
• find limits that contain $|x|$ and the greatest integer $[x]$;
• turn a limit at a point into a limit at $0$ by a substitution.

The squeeze theorem

Imagine a hallway with two walls. A person walks along it and the walls close in to the same spot. The person has to arrive at that spot. In symbols: suppose $g(x) \le f(x) \le h(x)$ for $x$ near $a$, and $\displaystyle\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$. Then $\displaystyle\lim_{x \to a} f(x) = L$.

Example 1. Find $\displaystyle\lim_{x \to 0} x\sin\dfrac{1}{x}$. The sine swings, but it is always between $-1$ and $1$. So $-|x| \le x\sin\dfrac{1}{x} \le |x|$. Both walls tend to $0$, so the limit is $0$.

Example 2. Find $\displaystyle\lim_{x \to \infty}\dfrac{\sin x}{x}$. We have $-\dfrac{1}{x} \le \dfrac{\sin x}{x} \le \dfrac{1}{x}$ for $x > 0$, so the limit is $0$.

Example 3. The standard limit $\dfrac{\sin x}{x} \to 1$ was squeezed too, between $\cos x$ and $1$.

The modulus

The modulus $|x|$ equals $x$ for $x \ge 0$ and $-x$ for $x < 0$. To take a limit, split into the left and the right.

Find $\displaystyle\lim_{x \to 1}\dfrac{|x - 1|}{x^2 - 1}$.

1. Right side, $x > 1$. $|x - 1| = x - 1$. The fraction is $\dfrac{1}{x + 1} \to \dfrac{1}{2}$.
2. Left side, $x < 1$. $|x - 1| = -(x - 1)$. The fraction is $-\dfrac{1}{x + 1} \to -\dfrac{1}{2}$.
3. Compare. The two one-sided limits differ, so the limit does not exist.

The greatest integer

The greatest integer $[x]$ is the largest integer that is not above $x$. So $[2.7] = 2$, $[3] = 3$ and $[-1.2] = -2$. It jumps at every integer. Some books write $\lfloor x \rfloor$. The fractional part is $\{x\} = x - [x]$.

• $\displaystyle\lim_{x \to 3^-}[x] = 2$ and $\displaystyle\lim_{x \to 3^+}[x] = 3$. The limit at $3$ does not exist.
• $\displaystyle\lim_{x \to 2.5}[x] = 2$, because $[x]$ is constant near $2.5$.

Use the squeeze. Since $[t] \le t < [t] + 1$, we get $t - 1 < [t] \le t$. Take $t = \dfrac{1}{x}$ with $x > 0$ and multiply by $x$: $1 - x < x\left[\dfrac{1}{x}\right] \le 1$. As $x \to 0^+$ both walls tend to $1$. So $\displaystyle\lim_{x \to 0^+} x\left[\dfrac{1}{x}\right] = 1$. For $x < 0$ the same argument gives $1$ again.

Similarly $\displaystyle\lim_{x \to \infty}\dfrac{[x]}{x} = 1$, because $\dfrac{x - 1}{x} < \dfrac{[x]}{x} \le 1$.

Turning $x \to a$ into $h \to 0$

Many trigonometric limits have $x \to a$ with $a \ne 0$. The standard limits need $0$. Put $x = a + h$, so that $h \to 0$.

Find $\displaystyle\lim_{x \to \pi/2}\dfrac{\cos x}{\dfrac{\pi}{2} - x}$.

1. Put $x = \dfrac{\pi}{2} - h$. Then $h \to 0$.
2. $\cos\left(\dfrac{\pi}{2} - h\right) = \sin h$. The fraction is $\dfrac{\sin h}{h}$.
3. The limit is $1$.

Another: $\displaystyle\lim_{x \to \pi}\dfrac{\sin x}{\pi - x}$. Put $x = \pi - h$ and use $\sin(\pi - h) = \sin h$. The limit is $1$.

A warning about oscillation at infinity

$\displaystyle\lim_{x \to \infty}\sin x$ does not exist, because the values keep swinging. But $\displaystyle\lim_{x \to \infty}\dfrac{x + \sin x}{x} = 1$, because $\dfrac{\sin x}{x} \to 0$. The bounded part loses against the growing part.

Practise with new numbers

Try it in Python

The program prints $x\sin\dfrac{1}{x}$ between its two walls for smaller and smaller $x$, and then $x\left[\dfrac{1}{x}\right]$.

Common mistakes

• Ignoring a side. For a modulus or $[x]$, check the left and the right.
• Squeezing with walls that do not meet. Both walls must tend to the same number.
• Treating $\sin\dfrac{1}{x}$ as small. On its own it does not tend to anything. Only the factor $x$ in front makes it vanish.
• Forgetting the negative side of $[x]$. $[-1.2] = -2$, not $-1$.

Remember these

• Squeeze: $g \le f \le h$ and $g, h \to L$ give $f \to L$.
• Split $|x - a|$ at $a$. Compare the one-sided limits.
• $t - 1 < [t] \le t$, which gives $x\left[\dfrac{1}{x}\right] \to 1$ as $x \to 0$.
• Put $x = a + h$ to turn $x \to a$ into $h \to 0$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find $\displaystyle\lim_{x \to 0} x^2\cos\dfrac{1}{x}$.

2. Find $\displaystyle\lim_{x \to 2^-}[x]$ and $\displaystyle\lim_{x \to 2^+}[x]$.

3. Find $\displaystyle\lim_{x \to \infty}\dfrac{3x + \sin x}{x + 2}$.

4. Find $\displaystyle\lim_{x \to 0^+}\dfrac{|x|}{x}$ and $\displaystyle\lim_{x \to 0^-}\dfrac{|x|}{x}$.

5. Find $\displaystyle\lim_{x \to \pi/2}\dfrac{1 - \sin x}{\dfrac{\pi}{2} - x}$.

References

• OpenStax, Calculus Volume 1, Chapter 2, section 2.3 on the squeeze theorem.
• NCERT, Mathematics Textbook for Class XI, Chapter 12, the sections on limits of trigonometric functions.