How a Function Can Head Towards 4 Without Ever Touching It

Take $f(x)=\frac{x^2-4}{x-2}$. At $x=2$ it gives $\frac{0}{0}$, so **$f(2)$ is not defined**. Yet close to $2$ the values behave very neatly:

- $x=1.9$ gives $3.9$ and $x=1.99$ gives $3.99$
- $x=2.1$ gives $4.1$ and $x=2.01$ gives $4.01$

The function **heads towards $4$** from both sides without ever being evaluated at $2$. That target value is the limit, and it is the foundation of all of calculus.

This part covers one-sided limits, the algebra of limits, 0/0 forms, and a standard result.

**${\lim }_{x\to a}f(x)=L$ means $f(x)$ gets as close to $L$ as we like when $x$ is close enough to $a$ but not equal to it; the limit exists only when the left-hand limit and the right-hand limit both exist and are equal.

- Left-hand limit** ${\lim }_{x\to a^{-}}f(x)$ — approach through values less than $a$
- Right-hand limit ${\lim }_{x\to a^{+}}f(x)$ — approach through values greater than $a$

Worked example 1. $f(x)=x^2$ for $x\le 1$ and $f(x)=2x-1$ for $x>1$.

$$\text{LHL}=1^2=1\text{RHL}=2(1)-1=1\Rightarrow \underset{x\to 1}{\lim }f(x)=1$$

Worked example 2. $f(x)=\frac{|x|}{x}$ at $x=0$ equals $-1$ for negative $x$ and $1$ for positive $x$, so LHL $=-1$ and RHL $=1$. The limit does not exist.

Worked example 3. The greatest integer function $[x]$ at $x=2$ has LHL $=1$ and RHL $=2$, so again no limit.

An everyday example. Suppose a courier charges ₹40 for parcels up to 500 g and ₹70 for anything heavier. Approaching 500 g from below gives ₹40, from above ₹70 — no single limiting charge.

The substance. **The limit never uses $f(a)$ itself** — $f(a)$ may be undefined or different from the limit.

**If ${\lim }_{x\to a}f(x)=l$ and ${\lim }_{x\to a}g(x)=m$, then the limit of a sum, difference, product or quotient is the sum, difference, product or quotient of the limits (with $m\ne 0$ for the quotient), so polynomials and rational functions with non-zero denominators are found by substitution.**

$$\lim (f\pm g)=l\pm m\lim (fg)=lm\lim \frac{f}{g}=\frac{l}{m}\lim (kf)=kl$$

Worked example 1 — polynomial.

$$\underset{x\to 2}{\lim }(3x^2-5x+4)=12-10+4=6$$

Worked example 2 — rational function.

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

Worked example 3 — product.

$$\underset{x\to 3}{\lim }(x+1)(x^2-2)=4\times 7=28$$

An everyday example. If the price of rice and the price of dal each settle to a steady value, the cost of buying both settles to the sum of those two values.

The substance. The rules need each separate limit to exist. When substitution makes the denominator zero, these rules alone cannot finish the job.

**A 0/0 result from substitution is not an answer but a signal: cancel the common factor $(x-a)$ by factorising, or multiply by the conjugate when square roots appear, and then substitute.

Worked example 1 — factorising.**

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

Worked example 2 — cubes.

$$\underset{x\to 2}{\lim }\frac{x^3-8}{x^2-4}=\underset{x\to 2}{\lim }\frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)}=\frac{4+4+4}{4}=3$$

Worked example 3 — rationalising.

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

Worked example 4.

$$\underset{x\to 4}{\lim }\frac{\sqrt{x}-2}{x-4}=\underset{x\to 4}{\lim }\frac{1}{\sqrt{x}+2}=\frac{1}{4}$$

An everyday example. The average speed of a bus over a shorter and shorter stretch of road divides a shrinking distance by a shrinking time, yet the ratio settles to a definite speed.

The substance. **Cancelling $(x-a)$ is allowed** because $x$ is never equal to $a$ while the limit is being taken.

**For a positive integer $n$, ${\lim }_{x\to a}\frac{x^n-a^n}{x-a}=n{a}^{n-1}$, because the quotient has exactly $n$ terms that each approach $a^{n-1}$; the result also holds for rational $n$ when $a>0$.

Why.**

$$x^n-a^n=(x-a)(x^{n-1}+x^{n-2}a+\cdots +a^{n-1})$$

Cancelling $(x-a)$ leaves $n$ terms, each tending to $a^{n-1}$.

Worked example 1.

$$\underset{x\to 2}{\lim }\frac{x^5-32}{x-2}=5\times 2^4=80$$

**Worked example 2 — divide top and bottom by $(x-1)$.**

$$\underset{x\to 1}{\lim }\frac{x^{15}-1}{x^{10}-1}=\frac{15\times 1^{14}}{10\times 1^9}=\frac{3}{2}$$

Worked example 3 — substitution. For ${\lim }_{x\to 0}\frac{(1+x{)}^6-1}{x}$, put $y=1+x$, so $y\to 1$:

$$\underset{y\to 1}{\lim }\frac{y^6-1}{y-1}=6$$

Worked example 4 — a fractional power. With $n=\frac{1}{2}$ and $a=4$, the result gives $\frac{1}{2}\times 4^{-1/2}=\frac{1}{4}$, matching the rationalising answer above.

An everyday example. Using this result is like a shopkeeper reading a ready-reckoner chart instead of multiplying out every bill — factorising $x^{15}-1$ by hand is not needed.

The substance. Match the form exactly — the same $a$ must appear in $x^n-a^n$ and in $x-a$.