Grade 11

Limits of functions

The value a function approaches as its input gets arbitrarily close to a point.

IntuitionWhat happens right next to a point

Suppose we cannot evaluate a formula right at $x=x_0$ — for instance the difference quotient $\frac{f(x_0+h)-f(x_0)}{h}$ at $h=0$ , which gives $\frac{0}{0}$ . Even though the point itself is off-limits, we can walk arbitrarily close to $x_0$ from both sides and watch what value $f(x)$ settles toward. That target value is the limit $\lim_{x\to x_0} f(x) = L$ , and it does not care whether $f(x_0)$ even exists.

Secant slope

\frac{f(x_0+h)-f(x_0)}{h}

: slide

h\to 0

and watch the secant line settle onto the tangent — a limit at

h=0

where the formula itself is undefined.

SchoolFormal definition and fundamental limits

Definition: Limit of a function at a point (ε–δ definition)

Let $f$ be defined on an open interval around $x_0$ , possibly excluding $x_0$ itself. We write $\lim_{x\to x_0} f(x) = L$ if for every tolerance $\varepsilon>0$ there is a radius $\delta>0$ such that $\forall \varepsilon>0,\ \exists \delta>0:\ 0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\varepsilon$ . Notice the strict inequality $0<|x-x_0|$ : $x$ is never allowed to equal $x_0$ in the test.

\forall \varepsilon>0,\ \exists \delta>0:\ 0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\varepsilon

Here $\varepsilon$ is the vertical tolerance around $L$ on the output axis, and $\delta$ (delta) is the horizontal window around $x_0$ on the input axis. Two foundational limits — proved by geometry and by monotone bounds — unlock all derivatives of trigonometric, exponential, and logarithmic functions: $\lim_{x\to 0}\frac{\sin x}{x}=1$ (with $x$ in radians) and $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e$ .

\lim_{x\to 0}\frac{\sin x}{x}=1,\qquad \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e

One-sided limits vs. two-sided limit at a point
Function and point	Left / right limits	Two-sided limit
$f(x)=\frac{\|x\|}{x}$ at $x_0=0$	Left $=-1$ , right $=+1$	Does not exist ( $-1\ne +1$ )
$f(x)=\frac{x^2-1}{x-1}$ at $x_0=1$	Left $=2$ , right $=2$	Equals $2$ (even though $f(1)$ is undefined)
$f(x)=\frac{\sin x}{x}$ at $x_0=0$	Left $=1$ , right $=1$	Equals $1$

UndergraduateKey theorems and proofs

Theorem: Existence of a limit via one-sided limits

A two-sided limit exists if and only if both one-sided limits exist and are equal: $\lim_{x\to x_0} f(x)=L \iff \lim_{x\to x_0^-} f(x)=\lim_{x\to x_0^+} f(x)=L$ .

Why is it true?

Walking toward a doorway from the left hallway and from the right hallway only lands you in the same room if both hallways actually meet at the same door.

Proof

( $\Rightarrow$ ) Suppose $\lim_{x\to x_0} f(x) = L$ . Given $\varepsilon>0$ , the ε–δ definition gives $\delta>0$ such that $0<|x-x_0|<\delta\Rightarrow |f(x)-L|<\varepsilon$ .

In particular, if $x_0<x<x_0+\delta$ (the right-hand condition) or $x_0-\delta<x<x_0$ (the left-hand condition), we automatically have $0<|x-x_0|<\delta$ , so $|f(x)-L|<\varepsilon$ . Thus both one-sided limits equal $L$ .

( $\Leftarrow$ ) Conversely, suppose both one-sided limits equal $L$ . For a given $\varepsilon>0$ , the left-hand limit gives $\delta_1>0$ working on $(x_0-\delta_1,x_0)$ , and the right-hand limit gives $\delta_2>0$ working on $(x_0,x_0+\delta_2)$ .

Set $\delta=\min(\delta_1,\delta_2)>0$ . Then any $x$ with $0<|x-x_0|<\delta$ lies either in $(x_0-\delta_1,x_0)$ or in $(x_0,x_0+\delta_2)$ , and in both cases $|f(x)-L|<\varepsilon$ . Hence $\lim_{x\to x_0} f(x) = L$ . $\blacksquare$

Theorem: Fundamental trigonometric limit

For $x$ measured in radians, $\lim_{x\to 0}\frac{\sin x}{x}=1$ .

Why is it true?

On a unit circle, a very small arc of length $x$ , its vertical chord leg $\sin x$ , and the outer tangent segment $\tan x$ become visually indistinguishable as the angle closes toward $0$ .

Proof

Take $0<x<\frac{\pi}{2}$ on the unit circle with center $O$ , point $A=(1,0)$ , point $B=(\cos x,\sin x)$ on the circle, and point $T=(1,\tan x)$ where the ray $OB$ meets the vertical tangent at $A$ .

The triangle $\triangle OAB$ sits strictly inside the circular sector $OAB$ , which sits strictly inside the right triangle $\triangle OAT$ . Comparing their areas gives $\frac{1}{2}\sin x < \frac{1}{2}x < \frac{1}{2}\tan x$ .

Since $\sin x>0$ on $(0,\frac{\pi}{2})$ , multiply through by $\frac{2}{\sin x}$ : $1 < \frac{x}{\sin x} < \frac{1}{\cos x}$ . Taking reciprocals reverses the inequalities: $\cos x < \frac{\sin x}{x} < 1$ .

As $x\to 0^+$ , $\cos x\to 1$ , so the Squeeze Theorem forces $\lim_{x\to 0^+}\frac{\sin x}{x}=1$ . Finally, $g(x)=\frac{\sin x}{x}$ is an even function ( $g(-x)=\frac{\sin(-x)}{-x}=\frac{-\sin x}{-x}=g(x)$ ), so $\lim_{x\to 0^-}\frac{\sin x}{x}=1$ as well, giving $\lim_{x\to 0}\frac{\sin x}{x}=1$ . $\blacksquare$

UndergraduateReal-World Applications and Worked Examples

Function limits turn average rates into instantaneous rates in physics (velocity, acceleration, electric current), justify the paraxial small-angle approximation $\sin\theta\approx\theta$ in geometric optics (camera lenses, telescopes) and pendulum mechanics, and turn discrete compounding into continuous compounding via $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e$ .

Example

A falling stone has position $s(t)=5t^2$ (in meters) at time $t$ (in seconds). Using a limit of average velocities over $[2,\ 2+h]$ , find its instantaneous velocity at $t=2$ s.

Solution

On the time interval $[2,\ 2+h]$ with $h\ne 0$ , the average velocity is $v_{\text{avg}}(h)=\frac{s(2+h)-s(2)}{h}=\frac{5(2+h)^2-5\cdot 2^2}{h}$ .

Expand the numerator: $5(4+4h+h^2)-20=20h+5h^2=h(20+5h)$ .

Since $h\ne 0$ in the limit, cancel $h$ : $v_{\text{avg}}(h)=20+5h$ .

Now take the limit as $h\to 0$ : $v(2)=\lim_{h\to 0}(20+5h)=20$ m/s. Notice how the $\frac{0}{0}$ indeterminate form disappears once we simplify before taking the limit.

Example

In paraxial optics, lens designers replace $\sin\theta$ in Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$ with $\theta$ (in radians) thanks to $\lim_{x\to 0}\frac{\sin x}{x}=1$ . Using the squeeze bound $\cos x < \frac{\sin x}{x} < 1$ , show that for an incident ray at $\theta=0.1$ rad ( $\approx 5.7^\circ$ ), the relative error $\frac{\theta-\sin\theta}{\sin\theta}$ is less than $1\%$ .

Solution

From $\cos x < \frac{\sin x}{x} < 1$ for $0<\theta<\frac{\pi}{2}$ , taking reciprocals gives $1<\frac{\theta}{\sin\theta}<\frac{1}{\cos\theta}$ .

Subtract $1$ across: $0<\frac{\theta-\sin\theta}{\sin\theta}<\frac{1}{\cos\theta}-1=\frac{1-\cos\theta}{\cos\theta}$ .

Using $1-\cos\theta=2\sin^2(\theta/2)<2(\theta/2)^2=\frac{\theta^2}{2}$ and $\cos(0.1)>0.99$ , we get at $\theta=0.1$ : $\frac{\theta-\sin\theta}{\sin\theta}<\frac{0.005}{0.99}\approx 0.00505$ , which is about $0.5\%<1\%$ .

(Indeed the exact value is $\approx 0.167\%$ .) This is why the linearized lens formula $\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$ is extremely accurate for rays close to the optical axis.

Common mistake. Wrong: "To find

\lim_{x\to x_0} f(x) = L

, just plug in

x=x_0

; if

f(x_0)

is undefined, the limit does not exist." Why it fails: the limit depends only on values of

f(x)

for

x\ne x_0

near

x_0

, never on

f(x_0)

itself — for example

\frac{x^2-1}{x-1}

is undefined at

x=1

, yet its limit as

x\to 1

2

. Also,

\lim_{x\to 0}\frac{\sin x}{x}=1

holds only when

x

is in radians; in degrees,

\lim_{x\to 0}\frac{\sin x^\circ}{x}=\frac{\pi}{180}

. Fix: simplify or apply a squeeze/conjugate identity on

x\ne x_0

before taking the limit, and always work in radians.

Compute $\lim_{x\to 2}\frac{x^2-4}{x-2}$ .

Let $f(x)=x+1$ for $x<1$ and $f(x)=x^2$ for $x\ge 1$ . What is $\lim_{x\to 1} f(x)$ ?

What is $\lim_{x\to 0}\frac{\sin(3x)}{x}$ (with $x$ in radians)?

Which famous constant is defined by $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x$ ?

References

James Stewart (2015). Calculus: Early Transcendentals
David Jerison (2010). MIT 18.01SC Single Variable Calculus, Session 4: Limits and Continuity