Math Doubts

$\displaystyle \lim_{x \to 0} \dfrac{\sin{x}}{x}$ Proof in Taylor/ Maclaurin Series Method

Take the literal $x$ as angle of the right angled triangle and the sine function is written as $\sin{x}$. the value of ratio of $\sin{x}$ to $x$ as the value of $x$ tends to $0$ is represented as the limit of ratio of $\sin{x}$ to $x$ when angle approaches zero in mathematical form.

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

It can be derived in calculus on the basis of Taylor expansion or Maclaurin series.

According to Taylor (or) Maclaurin Series, sine function can be expressed as an infinite series in terms of angle.

$\sin{x}$ $=$ $x$ $-\dfrac{x^3}{3!}$ $+$ $\dfrac{x^5}{5!}$ $-\dfrac{x^7}{7!}$ $\cdots$

$x$ is a common multiplying factor in each term of the expansion. So, take it common from them.

$\sin{x}$ $=$ $x\Bigg($ $1-\dfrac{x^2}{3!}$ $+$ $\dfrac{x^4}{5!}$ $-\dfrac{x^6}{7!} \cdots \Bigg)$

Now, express ratio of sin function to angle.

$\implies \dfrac{\sin{x}}{x}$ $= 1-\dfrac{x^2}{3!}$ $+$ $\dfrac{x^4}{5!}$ $-\dfrac{x^6}{7!} \cdots $

Now, calculate the value of ratio of sin function to angle as angle approaches $0$.

$\implies \large \displaystyle \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x}$ $= \large \displaystyle \lim_{x \,\to\, 0} \normalsize \Bigg(1-\dfrac{x^2}{3!}$ $+$ $\dfrac{x^4}{5!}$ $-\dfrac{x^6}{7!} \cdots \Bigg)$

$\implies \large \displaystyle \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x}$ $= 1-\dfrac{{(0)}^2}{3!}$ $+$ $\dfrac{{(0)}^4}{5!}$ $-\dfrac{{(0)}^6}{7!} \cdots $

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

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

On the basis of Taylor or Maclaurin series, it is proved that the value of limit of ratio of sin function to angle is equal to one as angle approaches zero.

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