What Happens to a Fraction When Its Top and Bottom Both Head to Zero

Speed at an instant, the slope of a curve at a point and the area under a curve all involve quantities that cannot be found by simply substituting a number — the formula breaks down exactly where you need it. Limits describe what a function approaches and make those ideas precise.

This lesson covers the idea of a limit, standard limits of different functions, indeterminate forms, and the theorems on limits.

**The limit of $f(x)$ as x approaches a is the value that $f(x)$ gets arbitrarily close to as x gets close to a from both sides, and it exists only when the left-hand limit and the right-hand limit are equal.

Notation:**

- Left-hand limit: ${\lim }_{x\to a^{-}}f(x)$
- Right-hand limit: ${\lim }_{x\to a^{+}}f(x)$
- ${\lim }_{x\to a}f(x)=L$ exactly when both one-sided limits equal L

Worked example. Let $f(x)=x+2$ for $x<1$ and $f(x)=4x-1$ for $x\ge 1$. From the left, $f(x)$ approaches $1+2=3$; from the right it approaches $4-1=3$. The limit exists and equals 3.

Worked example 2. For $f(x)=\frac{|x|}{x}$, the left-hand limit at 0 is $-1$ and the right-hand limit is 1, so ${\lim }_{x\to 0}f(x)$ does not exist.

Numerical view. For $f(x)=\frac{x^2-1}{x-1}$, the values at $x=0.99$ and $x=1.01$ are 1.99 and 2.01 — approaching 2, although $f(1)$ is undefined.

An everyday example. The reading on a car's speedometer is a limit: the average speed over shorter and shorter time intervals around one instant.

The substance. A limit can exist where the function is not defined — it depends on nearby values, not on $f(a)$ itself.

**A limit is found by direct substitution when the function is defined and well behaved at the point, and otherwise with standard results such as ${\lim }_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$, ${\lim }_{x\to 0}\frac{\sin x}{x}=1$, ${\lim }_{x\to 0}\frac{e^x-1}{x}=1$ and ${\lim }_{x\to 0}\frac{\log (1+x)}{x}=1$.

More standard limits** (x in radians, log to base e):

- ${\lim }_{x\to 0}\frac{\tan x}{x}=1$ and ${\lim }_{x\to 0}\frac{1-\cos x}{x}=0$
- ${\lim }_{x\to 0}\frac{a^x-1}{x}=\log a$

Worked examples:

- ${\lim }_{x\to 2}\frac{x^5-32}{x-2}=5\times 2^4=80$
- ${\lim }_{x\to 0}\frac{\sin 4x}{3x}=\frac{4}{3}{\lim }_{x\to 0}\frac{\sin 4x}{4x}=\frac{4}{3}$
- ${\lim }_{x\to 0}\frac{e^{3x}-1}{x}=3$
- ${\lim }_{x\to 0}\frac{2^x-1}{x}=\log 2\approx 0.693$

Numerical check. At $x=0.01$, $\frac{\sin 0.04}{0.03}\approx 1.333$.

An everyday example. Estimating the width of a distant hill from the small angle it makes at your eye uses $\sin x\approx x$ for small angles — the practical side of ${\lim }_{x\to 0}\frac{\sin x}{x}=1$.

The substance. The sine limit needs radians — measured in degrees, the ratio approaches $\frac{\pi }{180}$, not 1.

**When direct substitution gives a meaningless expression such as $\frac{0}{0}$ or $\frac{\infty }{\infty }$, the limit is found by factorising and cancelling, rationalising, or dividing by the highest power of x, and only then substituting.

Worked example (factorising).**

$$\underset{x\to 3}{\lim }\frac{x^2-9}{x^2-5x+6}=\underset{x\to 3}{\lim }\frac{(x-3)(x+3)}{(x-3)(x-2)}=\frac{6}{1}=6$$

Worked example (rationalising).

$$\underset{x\to 0}{\lim }\frac{\sqrt{1+x}-1}{x}=\underset{x\to 0}{\lim }\frac{1}{\sqrt{1+x}+1}=\frac{1}{2}$$

Worked example (large x). Dividing above and below by $x^2$:

$$\underset{x\to \infty }{\lim }\frac{3x^2+5}{2x^2-x}=\underset{x\to \infty }{\lim }\frac{3+\frac{5}{x^2}}{2-\frac{1}{x}}=\frac{3}{2}$$

An everyday example. **A courier's cost per parcel of $\frac{5000+40n}{n}$ rupees approaches ₹40 as the number of parcels n grows very large — a limit found by dividing by the highest power of n.

The substance. $\frac{0}{0}$ is a signal, not an answer** — it means more work is needed, and the limit may turn out to be any number.

**If ${\lim }_{x\to a}f(x)=L$ and ${\lim }_{x\to a}g(x)=M$, then the sum, difference and product have limits $L\pm M$ and $LM$, the quotient has limit $\frac{L}{M}$ when $M\ne 0$, and $kf(x)$ has limit $kL$.

The theorems:**

- $\lim [f(x)\pm g(x)]=L\pm M$
- $\lim [f(x)g(x)]=LM$
- $\lim \frac{f(x)}{g(x)}=\frac{L}{M}$, provided $M\ne 0$
- Sandwich theorem — if $g(x)\le f(x)\le h(x)$ and g and h have the same limit, then f has that limit too

Worked example. Evaluate ${\lim }_{x\to 0}\frac{\sin 2x+3x}{\tan x+4x}$. Dividing above and below by x:

$$\underset{x\to 0}{\lim }\frac{\frac{\sin 2x}{x}+3}{\frac{\tan x}{x}+4}=\frac{2+3}{1+4}=1$$

Worked example 2. ${\lim }_{x\to 0}{x}^2\sin \frac{1}{x}=0$ by the sandwich theorem, since $-x^2\le x^2\sin \frac{1}{x}\le x^2$ and both bounds tend to 0.

An everyday example. A cyclist whose speed approaches 12 km/h on the flat while a friend's approaches 8 km/h uphill has a speed difference approaching 4 km/h — limits combine the way ordinary numbers do.

The substance. The product theorem needs each separate limit to exist — $\sin \frac{1}{x}$ has no limit at 0, which is why the sandwich theorem is needed in the second example.