Limits

01The idea: approaching, not arriving

The statement $\lim\limits_{x \to a} f(x) = L$ means: as $x$ gets arbitrarily close to $a$ (from either side), the outputs $f(x)$ get arbitrarily close to $L$. The crucial subtlety is that the limit doesn't care what happens at $a$ itself — only what happens near $a$. The function can be undefined at $a$, or defined with some unrelated value, and the limit can still exist.

The hero graph above shows the classic example: $f(x) = \dfrac{x^2 - 9}{x - 3}$ is undefined at $x = 3$ (you'd get $\frac{0}{0}$), but everywhere else it simplifies to $x + 3$. So as $x$ approaches $3$, the outputs approach $6$. The graph is a straight line with a single hole punched out — and the limit is the $y$-value of the hole: $\lim\limits_{x\to 3} \dfrac{x^2-9}{x-3} = 6$.

02One-sided limits

Sometimes a function behaves differently on each side of a point. We write $\lim\limits_{x \to a^-} f(x)$ for the left-hand limit (approaching with $x < a$) and $\lim\limits_{x \to a^+} f(x)$ for the right-hand limit ($x > a$). The two-sided limit exists if and only if both one-sided limits exist and are equal:

$$\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L$$

This is the tool for piecewise functions and jump behavior. For example, $f(x) = \dfrac{|x|}{x}$ equals $-1$ for negative $x$ and $+1$ for positive $x$. The left limit at $0$ is $-1$, the right limit is $+1$ — they disagree, so $\lim\limits_{x\to 0}\dfrac{|x|}{x}$ does not exist (DNE).

03Limit laws and direct substitution

Limits behave politely with arithmetic: the limit of a sum is the sum of the limits, and the same holds for differences, products, and quotients (as long as the bottom limit isn't zero). The practical payoff is direct substitution: for polynomials, and for rational/trig/exponential/log functions at points inside their domains, you can just plug in:

$$\lim_{x \to 2}\,(x^2 + 3x - 1) = 4 + 6 - 1 = 9$$

Always try substitution first. If you get a real number, you're done. If you get $\frac{\text{nonzero}}{0}$, you're looking at a vertical asymptote (see §06). If you get $\frac{0}{0}$ — the indeterminate form — the answer is hiding and you need algebra.

04Defeating 0/0: factor, cancel, conjugate

An indeterminate form $\frac{0}{0}$ means numerator and denominator share a common factor that vanishes at $a$. Your job is to find it and cancel it. Two workhorse techniques:

1. Factor and cancel. $\lim\limits_{x\to 5}\dfrac{x^2 - 25}{x - 5} = \lim\limits_{x\to 5}\dfrac{(x-5)(x+5)}{x-5} = \lim\limits_{x\to 5}(x+5) = 10$.
2. Multiply by the conjugate when square roots appear. For $\lim\limits_{x\to 0}\dfrac{\sqrt{x+4}-2}{x}$, multiply top and bottom by $\sqrt{x+4}+2$: the numerator becomes $(x+4)-4 = x$, which cancels the denominator, leaving $\lim\limits_{x\to 0}\dfrac{1}{\sqrt{x+4}+2} = \dfrac{1}{4}$.

After canceling, substitute again. This is why the factoring drills from Module 01 matter — they are the limit technique.

05The Squeeze Theorem

If $g(x) \le f(x) \le h(x)$ near $a$, and the outer functions squeeze together — $\lim\limits_{x\to a} g(x) = \lim\limits_{x\to a} h(x) = L$ — then the trapped function must go there too: $\lim\limits_{x\to a} f(x) = L$. The signature application: since $-1 \le \sin\frac{1}{x} \le 1$, we get $-|x| \le x\sin\frac{1}{x} \le |x|$, and both bounds go to $0$, so $\lim\limits_{x\to 0} x\sin\frac{1}{x} = 0$ — even though the function oscillates infinitely fast near zero.

06Infinite limits and vertical asymptotes

When substitution gives $\frac{\text{nonzero}}{0}$, the function blows up. We write $\lim\limits_{x\to a} f(x) = \infty$ (or $-\infty$) to describe how it fails to exist, and the line $x = a$ is a vertical asymptote. Determine the sign by checking each side: for $\lim\limits_{x\to 2^+}\dfrac{1}{x-2}$, values just right of $2$ make $x - 2$ a tiny positive number, so the quotient is huge and positive: $+\infty$. From the left it's $-\infty$.

07Limits at infinity and horizontal asymptotes

Now let $x$ itself run off to $\pm\infty$. For rational functions, compare the degrees of the top and bottom:

• Top degree < bottom degree — limit is $0$.
• Degrees equal — limit is the ratio of leading coefficients.
• Top degree > bottom degree — limit is $\pm\infty$ (no horizontal asymptote).

So $\lim\limits_{x\to\infty} \dfrac{3x^2 + x}{5x^2 - 7} = \dfrac{3}{5}$, and $y = \frac{3}{5}$ is a horizontal asymptote. The formal move behind this shortcut: divide top and bottom by the highest power of $x$ in the denominator, and use $\lim\limits_{x\to\infty}\frac{1}{x^n} = 0$.

08Two special trig limits

Memorize these — they power the derivatives of sine and cosine later:

$$\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad\qquad \lim_{x \to 0} \frac{1 - \cos x}{x} = 0$$

(Radians required — another reason degrees are banned in calculus.) These stretch further than they look: $\lim\limits_{x\to 0}\dfrac{\sin 5x}{x} = \lim\limits_{x\to 0}\dfrac{\sin 5x}{5x}\cdot 5 = 1 \cdot 5 = 5$. The pattern: match the inside of the sine with the denominator, then account for the constant you introduced.