$\displaystyle \lim_{x \to 0} \dfrac{\sin{x}}{x}$ Proof in General Method

$x$ is a literal but represents angle of a right angled triangle and the sine function is represented by $\sin{x}$.

The ratio of $\sin{x}$ to $x$ is expressed as $\dfrac{\sin{x}}{x}$. The value of ratio of $\sin{x}$ to $x$ as $x$ approaches $0$ is written in the following mathematical form.

$\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x}$

As the angle tends to zero, a special close relation between angle and sine of angle is revealed. In this special case, the value of sine function is approximately equal to the angle. Observe the following table to understand it.

Angle $(x)$ in Radian $\sin{x}$
$0$ $0$
$0.1$ $0.099833417$ $\approx 0.1$
$0.02$ $0.019998667$ $\approx 0.02$
$0.003$ $0.002999996$ $\approx 0.03$
$0.0004$ $0.0004$
$0.00005$ $0.00005$
$0.000123$ $0.000123$
$0.0006583$ $0.0006583$
$0.00027595$ $0.00027595$
$0.000039178$ $0.000039178$

Therefore, $\sin{x} \approx x$ as $x$ approaches zero.

$\implies \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x} = \large \lim_{x \,\to\, 0} \normalsize \dfrac{x}{x}$

$\require{cancel} \implies \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x} = \large \lim_{x \,\to\, 0} \normalsize \dfrac{\cancel{x}}{\cancel{x}}$

$\implies \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x} = \large \lim_{x \,\to\, 0} \normalsize {1}$

$\,\,\, \therefore \,\,\,\,\,\, \displaystyle \large \lim_{x \,\to\, 0} \dfrac{\sin{x}}{x} = 1$

Therefore, it is proved that the value of ratio of sin function to angle is equal to one as the angle approaches zero.

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