Limits of Trigonometric Functions

The Sandwich Theorem and $\lim_{x \to 0} \frac{\sin x}{x} = 1$

Two comparison facts drive every trigonometric limit.

Key Point (Theorems 3-4): if $f(x) \le g(x)$ on a common domain and both limits exist at $a$, then $\lim f \le \lim g$. And the Sandwich Theorem: if $f(x) \le g(x) \le h(x)$ and $\lim_{x \to a} f(x) = l = \lim_{x \to a} h(x)$, then $\lim_{x \to a} g(x) = l$ — the middle function has nowhere else to go.

The bread for the sandwich comes from areas in a unit circle. For $0 < x < \frac{\pi}{2}$, comparing triangle $OAC$, sector $OAC$ and triangle $OAB$:

$$\sin x < x < \tan x \;\Rightarrow\; \cos x < \frac{\sin x}{x} < 1$$

(dividing by $\sin x > 0$ and taking reciprocals). Since $\cos x \to 1$, the sandwich closes:

Key Point (Theorem 5):

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

The second follows from the first via $1 - \cos x = 2\sin^2\frac{x}{2}$. Both need $x$ in radians — the area comparison lives on the unit circle.

The Scaling Toolkit

Every trigonometric limit in this chapter reduces to copies of $\frac{\sin\theta}{\theta}$ with $\theta \to 0$:

Scaling: $\lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b}$ — multiply and divide by $ax$: $\frac{\sin ax}{ax} \cdot \frac{a}{b} \to 1 \cdot \frac{a}{b}$. Similarly $\lim_{x \to 0}\frac{\sin ax}{\sin bx} = \frac{a}{b}$ — e.g. $\lim_{x \to 0} \frac{\sin 4x}{\sin 2x} = 2$.

tan follows sin: $\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x} \to 1$.

$1 - \cos$ pairs with half-angles: $\frac{1 - \cos 2x}{1 - \cos x} = \frac{2\sin^2 x}{2\sin^2\frac{x}{2}} \to \frac{x^2}{x^2/4} = 4$ (using $\sin\theta \approx \theta$ through the fundamental limit).

Shifts: near $a \ne 0$, substitute $y = x - a$ so the new variable tends to 0. For example $\lim_{x \to \pi/2}\frac{\tan 2x}{x - \frac{\pi}{2}}$ becomes, with $y = x - \frac{\pi}{2}$ and $\tan(2y + \pi) = \tan 2y$, the limit $\frac{\tan 2y}{y} \to 2$.

[Board Tip] Before invoking $\frac{\sin\theta}{\theta} \to 1$, make sure the same expression sits in both numerator and denominator: $\frac{\sin 4x}{x}$ is $4 \cdot \frac{\sin 4x}{4x} \to 4$, not 1. The multiply-and-divide step is where the marks live.

Solved Examples

Example 1: The scaling template

Evaluate: (i) $\lim_{x \to 0} \frac{\sin 4x}{\sin 2x}$; (ii) $\lim_{x \to 0} \frac{\tan x}{x}$.

Solution:

Step 1 — (i) Scale each sine by its own argument. $\frac{\sin 4x}{\sin 2x} = \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4x}{2x}$.

Step 2 — (i) Pass to the limit. The two scaled sines $\to 1$, leaving $\frac{4}{2} = 2$.

Step 3 — (ii) Split tan. $\frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x} \to 1 \times 1 = 1$.

Takeaway: Multiply-and-divide so every sine sits over its OWN argument — the ratio of the arguments is the answer.

Example 2: General scalings

Evaluate: (i) $\lim_{x \to 0} \frac{\sin ax}{bx}$; (ii) $\lim_{x \to 0} \frac{\sin ax}{\sin bx}$ ($a, b \ne 0$).

Solution:

Step 1 — (i) Insert $ax$. $\frac{\sin ax}{bx} = \frac{\sin ax}{ax} \cdot \frac{a}{b}$.

Step 2 — (i) Limit. $1 \cdot \frac{a}{b} = \frac{a}{b}$.

Step 3 — (ii) Same idea twice. $\frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{a}{b} \to \frac{a}{b}$.

Takeaway: $\frac{\sin(\text{fast})}{\text{slow}}$ scales by the ratio of the rates — memorise $\frac{a}{b}$ and re-derive when unsure.

Example 3: Shifts and quiet denominators

Evaluate: (i) $\lim_{x \to \pi} \frac{\sin(\pi - x)}{\pi(\pi - x)}$; (ii) $\lim_{x \to 0} \frac{\cos x}{\pi - x}$.

Solution:

Step 1 — (i) Substitute the vanishing quantity. Let $y = \pi - x$; as $x \to \pi$, $y \to 0$.

Step 2 — (i) Recognise the fundamental limit. $\frac{1}{\pi} \cdot \frac{\sin y}{y} \to \frac{1}{\pi}$.

Step 3 — (ii) Check before machinery. At 0 nothing vanishes in the denominator: substitute directly, $\frac{\cos 0}{\pi - 0} = \frac{1}{\pi}$.

Takeaway: First ask WHETHER anything tends to 0 — part (ii) needs no trick at all, and spotting that saves a minute.

Example 4: Double angles

Evaluate $\lim_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}$.

Solution:

Step 1 — Half-angle both. $\cos 2x - 1 = -2\sin^2 x$; $\cos x - 1 = -2\sin^2\frac{x}{2}$.

Step 2 — Form the ratio. $\frac{\sin^2 x}{\sin^2\frac{x}{2}}$ (the minus signs cancel).

Step 3 — Scale each sine. $\left(\frac{\sin x}{x}\right)^2 x^2$ over $\left(\frac{\sin(x/2)}{x/2}\right)^2 \frac{x^2}{4}$ tends to $\frac{x^2}{x^2/4} = 4$.

Takeaway: $1 - \cos(\text{anything})$ becomes $2\sin^2(\text{half})$ — after that it is pure scaling.

Example 5: Mixed algebraic-trig

Evaluate: (i) $\lim_{x \to 0} \frac{ax + x\cos x}{b\sin x}$; (ii) $\lim_{x \to 0} x \sec x$.

Solution:

Step 1 — (i) Factor the common $x$. $\frac{x(a + \cos x)}{b \sin x} = \frac{a + \cos x}{b} \cdot \frac{x}{\sin x}$.

Step 2 — (i) Limit each factor. $\frac{a + 1}{b} \times 1 = \frac{a+1}{b}$.

Step 3 — (ii) No indeterminacy. $x \sec x = \frac{x}{\cos x} \to \frac{0}{1} = 0$.

Takeaway: $\frac{x}{\sin x} \to 1$ works in both directions — the fundamental limit is symmetric in numerator and denominator.

Example 6: sin and linear terms together

Evaluate $\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$ ($a, b, a + b \ne 0$).

Solution:

Step 1 — Divide top and bottom by $x$. $\frac{a \cdot \frac{\sin ax}{ax} + b}{a + b \cdot \frac{\sin bx}{bx}}$.

Step 2 — Pass to the limit. Each scaled sine $\to 1$: $\frac{a + b}{a + b}$.

Step 3 — Conclude. The limit is 1 — with the hypothesis $a + b \ne 0$ making the final division legal.

Takeaway: Dividing by $x$ converts every term into either a constant or a scaled sine — the universal opener for sin-plus-linear ratios.

Example 7: cosec minus cot

Evaluate $\lim_{x \to 0} (\mathrm{cosec}\, x - \cot x)$.

Solution:

Step 1 — Combine over $\sin x$. $\mathrm{cosec}\, x - \cot x = \frac{1 - \cos x}{\sin x}$.

Step 2 — Half-angle everything. $\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = \tan\frac{x}{2}$.

Step 3 — Limit. $\tan\frac{x}{2} \to \tan 0 = 0$.

Takeaway: An $\infty - \infty$ of trig functions usually collapses under a common denominator plus half-angle identities — never limit the pieces separately.

Example 8: A shift to $\frac{\pi}{2}$

Evaluate $\lim_{x \to \pi/2} \frac{\tan 2x}{x - \frac{\pi}{2}}$.

Solution:

Step 1 — Shift the variable. Let $y = x - \frac{\pi}{2} \to 0$; then $2x = 2y + \pi$.

Step 2 — Use periodicity. $\tan(2y + \pi) = \tan 2y$.

Step 3 — Scale. $\frac{\tan 2y}{y} = 2 \cdot \frac{\tan 2y}{2y} \to 2 \times 1 = 2$.

Takeaway: Shift so the new variable tends to 0, then let periodicity clean up the trig — the $\pi$ vanished by itself.

Example 9: The second fundamental limit at work

Evaluate $\lim_{x \to 0} \frac{1 - \cos x}{x^2}$.

Solution:

Step 1 — Half-angle. $\frac{1 - \cos x}{x^2} = \frac{2\sin^2\frac{x}{2}}{x^2}$.

Step 2 — Scale the sine. $= \frac{1}{2}\left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \to \frac{1}{2}$.

Step 3 — Compare with the $x$ version. Against $x$ the numerator gives 0; against $x^2$ it gives $\frac{1}{2}$ — so $1 - \cos x$ behaves like $\frac{x^2}{2}$ near 0.

Takeaway: $\frac{1-\cos x}{x} \to 0$ but $\frac{1-\cos x}{x^2} \to \frac{1}{2}$ — knowing the ORDER of vanishing is what JEE tests.

Example 10: A sin-tan mix

Evaluate $\lim_{x \to 0} \frac{\sin 3x + 7x}{4x + \sin 2x}$.

Solution:

Step 1 — Divide by $x$. $\frac{3\frac{\sin 3x}{3x} + 7}{4 + 2\frac{\sin 2x}{2x}}$.

Step 2 — Limit each piece. $\frac{3(1) + 7}{4 + 2(1)} = \frac{10}{6}$.

Step 3 — Simplify. $\frac{5}{3}$.

Takeaway: Every $\sin kx$ contributes its coefficient $k$ after dividing by $x$ — read the answer as $\frac{3 + 7}{4 + 2}$ at sight.

Example 11: A sandwich argument

Show that $\lim_{x \to 0} x^2 \sin\frac{1}{x} = 0$.

Solution:

Step 1 — Bound the oscillation. $\left|\sin\frac{1}{x}\right| \le 1$ for every $x \ne 0$, so $-x^2 \le x^2 \sin\frac{1}{x} \le x^2$.

Step 2 — Squeeze. Both outer functions tend to 0 as $x \to 0$.

Step 3 — Conclude. By the Sandwich Theorem the middle limit is 0. ∎

Takeaway: "Bounded × vanishing → 0" — the sandwich handles wild oscillation that no substitution could.

Example 12: Radians matter

What is $\lim_{x \to 0} \frac{\sin x°}{x}$, where $x°$ means $x$ degrees?

Solution:

Step 1 — Convert. $x° = \frac{\pi x}{180}$ radians.

Step 2 — Rebuild the fundamental ratio. $\frac{\sin x°}{x} = \frac{\pi}{180} \cdot \frac{\sin\frac{\pi x}{180}}{\frac{\pi x}{180}}$.

Step 3 — Limit. $\frac{\pi}{180} \times 1 = \frac{\pi}{180}$ — not 1.

Takeaway: $\frac{\sin\theta}{\theta} \to 1$ is a RADIAN theorem — the unit-circle area proof breaks in degrees, and the honest answer carries the conversion factor.