Evaluate the limit of sin(x)/x as x approaches 0.

Step-by-step solution

1. 1

   Check the form. As $x\to0$ both $\sin x\to0$ and $x\to0$, so substitution gives $\tfrac00$ — indeterminate. This limit cannot be evaluated by plugging in, and it cannot honestly be evaluated by L'Hopital either, because the derivative $\frac{d}{dx}\sin x=\cos x$ is itself normally proved using this very limit.

2. 2

   Set up the geometric comparison. For $0<x<\tfrac{\pi}{2}$, compare three areas inside the unit circle: the triangle with area $\tfrac12\sin x$, the sector with area $\tfrac12x$, and the larger triangle with area $\tfrac12\tan x$. Since each region contains the previous one,

   $\frac12\sin x\le\frac12x\le\frac12\tan x.$

   The middle term is where the radian measure enters: the sector area is $\tfrac12r^{2}x$ only when $x$ is in radians.

3. 3

   Rearrange into a squeeze. Dividing throughout by $\tfrac12\sin x>0$ gives $1\le\frac{x}{\sin x}\le\frac{1}{\cos x}$, and taking reciprocals reverses the inequalities:

   $\cos x\le\frac{\sin x}{x}\le1.$

   Both $\frac{\sin x}{x}$ and $\cos x$ are even functions, so the same bounds hold for $-\tfrac{\pi}{2}<x<0$ and the argument covers both sides at once.

4. 4

   Apply the squeeze theorem. Since $\cos x\to1$ and the constant $1\to1$ as $x\to0$, the trapped quantity has no choice:

   $\lim_{x\to0}\frac{\sin x}{x}=1.$

5. 5

   Confirm numerically and note the radian caveat. Evaluating in radians: at $x=0.1$ the ratio is $0.998334$, at $x=0.01$ it is $0.9999833$, at $x=0.001$ it is $0.99999983$ — approaching $1$ from below, matching the bound $\cos x\le\frac{\sin x}{x}$ ✓. If $x$ were measured in degrees the limit would instead be $\frac{\pi}{180}\approx0.01745$, which is precisely why calculus is done in radians.