In this limit problem, an algebraic function in square form and a trigonometric function in cosine form are defined in a variable $x$, and the two functions formed the following rational expression.

$\dfrac{1-\cos{(2x)}}{x^2}$

We have to evaluate the limit of this function as $x$ approaches $0$ in this trigonometric limit problem.

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

Let’s try the fundamental method for evaluating the limit of the function as $x$ tends to $0$.

$=\,\,\,$ $\dfrac{1-\cos{(2(0))}}{(0)^2}$

$=\,\,\,$ $\dfrac{1-\cos{(2 \times 0)}}{0^2}$

$=\,\,\,$ $\dfrac{1-\cos{(0)}}{0}$

According to trigonometry, the cos of zero degrees is one.

$=\,\,\,$ $\dfrac{1-1}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

The limit of the quotient of $1-\cos{(2x)}$ by $x^2$ is indeterminate, which expresses that the direct substitution method is failed for evaluating the limit of the function. Hence, we have to think for another mathematical approach instead of direct substitution.

In numerator of the rational expression, there are two terms but there is only one term in the denominator, which clears that the expression in the numerator can be simplified and no need to focus on the expression in the denominator.

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

We cannot expand the first term but we can expand the second term $\cos{2x}$ by the cos double angle identity. As per cos double angle formula, the term $\cos{2x}$ can be expanded in cosine or sine but there is no limit rule in cosine. Hence, it is better to expand the cosine double angle function in sine.

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

Now, let’s focus on simplifying the mathematical expression.

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

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

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

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

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

We can use the constant multiple rule of limits for taking the constant out from the limit operation.

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

The quotient of square functions can be simplified by the power rule of quotient.

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

Now, use the power rule of limits for simplifying the mathematical expression further.

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

Now, use the limit rule of sinx/x as x approaches 0 for evaluating the limit of the simplified function.

$=\,\,\,$ $2 \times (1)^2$

$=\,\,\,$ $2 \times 1^2$

$=\,\,\,$ $2 \times 1$

$=\,\,\,$ $2$

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