According to the direct substitution method, the limit of one minus cosine of angle x by x square as x approaches 0 is indeterminate. Hence, let’s try to find the limit of the $1-\cos{x}$ by $x^2$ as $x$ closer to $0$ by using the combination of both trigonometric identities and limit rules.

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{x}}{x^2}}$

Look at the trigonometric expression in the numerator and it is in terms of cosine function but there is no trigonometric limit rule in terms of cosine. However, the cosine function can be converted into sine function by using the half angle cosine identity.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\bigg(1-2\sin^2{\Big(\dfrac{x}{2}\Big)}\bigg)}{x^2}}$

Now, simplify the trigonometric expression in the numerator.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-1+2\sin^2{\Big(\dfrac{x}{2}\Big)}}{x^2}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cancel{1}-\cancel{1}+2\sin^2{\Big(\dfrac{x}{2}\Big)}}{x^2}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{x^2}}$

The simplified mathematical expression is similar to the trigonometric limit rule in sine function. So, let’s try to transform the mathematical expression into required limit rule’s form.

Suppose $y \,=\, \dfrac{x}{2}$ for our convenience. Then, $x \,=\, 2y$. Now, transform the mathematical expression in terms of $y$ by using the mathematical relations between $x$ and $y$.

$\implies$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{x}{2}\Big)}}{x^2}}$ $\,=\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \dfrac{2\sin^2{y}}{(2y)^2}}$

Now, let us focus on converting the mathematical expression in terms of $y$ into our requisite form.

$=\,\,\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \dfrac{2\sin^2{y}}{(2 \times y)^2}}$

The power of a product rule can be used to express the square of the product of $2$ and $y$ as a product of their squares.

$=\,\,\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \dfrac{2\sin^2{y}}{2^2 \times y^2}}$

$=\,\,\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \dfrac{2\sin^2{y}}{4 \times y^2}}$

It is time to eliminate the constant factors from the limit operation by expressing the function as a product of two fractional functions.

$=\,\,\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \dfrac{2 \times \sin^2{y}}{4 \times y^2}}$

$=\,\,\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \bigg(\dfrac{2}{4} \times \dfrac{\sin^2{y}}{y^2}\bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \bigg(\dfrac{\cancel{2}}{\cancel{4}} \times \dfrac{\sin^2{y}}{y^2}\bigg)}$

$=\,\,\,$ $\displaystyle \large \lim_{2y\,\to\,0}{\normalsize \bigg(\dfrac{1}{2} \times \dfrac{\sin^2{y}}{y^2}\bigg)}$

According to the constant multiple rule of limits, the constant factor has an exception from the limiting operation.

$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{2y\,\to\,0}{\normalsize \dfrac{\sin^2{y}}{y^2}}$

The angle in the sine function is $y$ and it is also there in denominator but the input of the limit operation is double. It indicates that this expression requires further simplification to apply the trigonometric limit rule in sine function.

If $2y \,\to\, 0$, then $\dfrac{2y}{2} \,\to\, \dfrac{0}{2}$. It means $\dfrac{\cancel{2}y}{\cancel{2}} \,\to\, 0$. Therefore, $y\,\to\, 0$.

It clears that the value of $y$ also approaches $0$ when the value of $2y$ is closer to $0$.

$\implies$ $\dfrac{1}{2} \times \displaystyle \large \lim_{2y\,\to\,0}{\normalsize \dfrac{\sin^2{y}}{y^2}}$ $\,=\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin^2{y}}{y^2}}$

Now, let’s continue the procedure of converting the mathematical expression into the required form.

$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{(\sin{y})^2}{y^2}}$

The quotient of the sine squared of angle $y$ by $y$ squared can be written as the square of their quotient as per the power of a quotient rule.

$=\,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{y\,\to\,0}{\normalsize \bigg(\dfrac{\sin{y}}{y}\bigg)^2}$

Now, we can use the power rule of limits to get the required form.

$=\,\,\,$ $\dfrac{1}{2} \times \bigg(\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{y}}{y}\bigg)^2}$

The given mathematical expression is successfully converted into the required form as follows.

$=\,\,\,$ $\dfrac{1}{2} \times \bigg(\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{y}}{y}\bigg)^2}$

As per the trigonometric limit rule in sine function, the limit of the sine of angle $y$ by $y$ as $y$ tends to $0$ is equal to one.

$=\,\,\,$ $\dfrac{1}{2} \times (1)^2$

$=\,\,\,$ $\dfrac{1}{2} \times 1$

$=\,\,\,$ $\dfrac{1}{2}$

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