Prove that lim(x→0) sin x/x = 1 using squeeze theorem.

Find lim(x→0) sin x/x — Fundamental Limit

Question

Prove that:

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

where $x$ is measured in radians.

Solution — Step by Step

If we substitute $x = 0$ directly, we get $\frac{\sin 0}{0} = \frac{0}{0}$ — an indeterminate form. This tells us the limit might still exist, but we need a geometric argument to find it.

Consider a unit circle (radius = 1). For a small positive angle $x$ radians, we compare three areas:

Triangle OAP (inscribed): area $= \frac{1}{2} \sin x$
Circular sector OAP: area $= \frac{1}{2} x$
Triangle OAT (circumscribed): area $= \frac{1}{2} \tan x$

Since the triangle fits inside the sector, which fits inside the outer triangle:

\frac{1}{2}\sin x \leq \frac{1}{2}x \leq \frac{1}{2}\tan x

Dividing all parts by $\frac{1}{2}\sin x$ (positive for small $x > 0$ ):

1 \leq \frac{x}{\sin x} \leq \frac{1}{\cos x}

Taking reciprocals flips the inequalities:

\cos x \leq \frac{\sin x}{x} \leq 1

As $x \to 0$ , we know $\cos x \to 1$ . So our expression $\frac{\sin x}{x}$ is squeezed between $\cos x$ and $1$ , both approaching $1$ .

By the Squeeze Theorem (Sandwich Theorem):

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

The argument above works for $x > 0$ . For $x < 0$ , let $x = -t$ where $t > 0$ :

\frac{\sin x}{x} = \frac{\sin(-t)}{-t} = \frac{-\sin t}{-t} = \frac{\sin t}{t}

Since this equals the positive case, the limit from both sides is $1$ . Hence the two-sided limit equals $\boxed{1}$ .

Why This Works

The Squeeze Theorem is the right tool here because we cannot algebraically simplify $\frac{\sin x}{x}$ — there’s no cancellation. Instead, we trap the function between two simpler functions that converge to the same value.

The geometric setup on the unit circle is the key insight. Areas give us a natural inequality because containment of regions directly translates into containment of their areas. The sector area $\frac{1}{2}x$ uses the fact that for a unit circle, arc length equals the angle in radians — which is precisely why this whole argument requires radian measure.

If you try this limit with $x$ in degrees, you get $\frac{\pi}{180}$ , not $1$ . This is why calculus always works in radians — derivatives of trig functions come out clean only then.

Alternative Method

For JEE, you’ll often use the result directly as a standard limit. But L’Hôpital’s Rule also works (though it’s circular if you’re trying to derive the derivative of $\sin x$ ):

Since $\frac{\sin x}{x}$ is $\frac{0}{0}$ form at $x = 0$ , differentiate numerator and denominator separately:

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

In JEE Main and board exams, this result is a direct standard limit — you’re allowed to quote it without proof. What gets asked are problems that reduce to this form, like $\lim_{x \to 0} \frac{\sin 3x}{x}$ or $\lim_{x \to 0} \frac{\tan x}{x}$ .

Common Mistake

Forgetting to check radian measure. Students apply $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to expressions where $x$ is implicitly in degrees. If the problem involves a degree-based formula, the answer is $\frac{\pi}{180}$ , not $1$ . Always confirm you’re working in radians before using this standard limit.

A second common slip: flipping the inequalities incorrectly in Step 3. When you take reciprocals of a chain $a \leq b \leq c$ (all positive), it becomes $\frac{1}{c} \leq \frac{1}{b} \leq \frac{1}{a}$ — the order reverses. Missing this flip gives the wrong squeeze direction.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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