Evaluate limit as x→0 of (sinx)/x using squeeze theorem

Question

Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin x}{x}$ using the squeeze theorem.

Solution — Step by Step

Consider a unit circle (radius = 1) with a small positive angle $x$ (in radians). Three regions can be compared by area:

Triangle OAP (small): area $= \frac{1}{2} \sin x$
Sector OAP: area $= \frac{1}{2} x$ (for a unit circle, sector area $= \frac{1}{2}r^2\theta = \frac{x}{2}$ )
Triangle OAT (large): area $= \frac{1}{2} \tan x$

Since small triangle $\subset$ sector $\subset$ large triangle:

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

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

Take reciprocals (reversing inequalities):

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

As $x \to 0$ :

\lim_{x \to 0} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0} 1 = 1

Since $\cos x \leq \dfrac{\sin x}{x} \leq 1$ and both outer functions approach 1:

By the Squeeze Theorem (Sandwich Theorem):

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

For $x < 0$ : let $x = -t$ where $t > 0$ :

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

So the left-hand limit also equals 1. Therefore:

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

Why This Works

The squeeze theorem says: if $f(x) \leq g(x) \leq h(x)$ near $x = a$ , and $\lim f(x) = \lim h(x) = L$ , then $\lim g(x) = L$ too. The function $\frac{\sin x}{x}$ is “squeezed” between $\cos x$ and 1, both of which approach 1 as $x \to 0$ , forcing $\frac{\sin x}{x}$ to approach 1 as well.

This result is fundamental — it’s used to derive that the derivative of $\sin x$ is $\cos x$ , which in turn underlies all of trigonometric calculus.

Alternative Method — L’Hôpital’s Rule (Not Rigorous for Proving This)

Technically, applying L’Hôpital to $\frac{\sin x}{x}$ gives $\frac{\cos x}{1} \to 1$ . But this is circular: deriving $(\sin x)' = \cos x$ requires knowing $\lim \frac{\sin x}{x} = 1$ in the first place. The squeeze theorem proof is the proper one.

Common Mistake

The limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$ holds only when $x$ is in radians. If $x$ is in degrees, $\frac{\sin x°}{x} \to \frac{\pi}{180} \neq 1$ . In calculus, angles are always in radians unless specified otherwise.

This standard limit appears in JEE almost every year in slightly disguised forms: $\lim_{x\to 0} \frac{\sin 3x}{x}$ , $\lim_{x\to 0} \frac{\tan x}{x}$ , $\lim_{x\to 0} \frac{1-\cos x}{x^2}$ . The key: substitute $t = 3x$ (or similar) to reduce to the standard form. For example, $\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x} \to 3 \times 1 = 3$ .

Question

Solution — Step by Step

Why This Works

Alternative Method — L’Hôpital’s Rule (Not Rigorous for Proving This)

Common Mistake

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