Math Doubts

Evaluate $\displaystyle \large \lim_{x \,\to\, \frac{\pi}{2}}{\normalsize \dfrac{\cos{x}}{\frac{\pi}{2}-x}}$

In this calculus problem, the limit of the quotient of $\cos{x}$ by $\dfrac{\pi}{2}-x$ should have to evaluate as $x$ approaches $\dfrac{\pi}{2}$. In this problem, the $x$ represents a variable and also represents an angle of a right triangle.

$\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{\cos{x}}{\dfrac{\pi}{2}-x}}$

Now. let’s try to evaluate the limit of the algebraic trigonometric function as $x$ approaches $\dfrac{\pi}{2}$ by using direct substitution method.

$= \,\,\,$ $\dfrac{\cos{\Big(\dfrac{\pi}{2}\Big)} }{\dfrac{\pi}{2}-\dfrac{\pi}{2}}$

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

It is evaluated that the limit of the given function is indeterminate as the input $x$ approaches $\dfrac{\pi}{2}$. So, it should be evaluated in another method.

Express the function in another variable

Take $y = \dfrac{\pi}{2}-x$, then $x = \dfrac{\pi}{2}-y$.

$(1) \,\,\,$ If $x \to \dfrac{\pi}{2}$ then $x-\dfrac{\pi}{2} \to \dfrac{\pi}{2}-\dfrac{\pi}{2}$. Therefore, $x-\dfrac{\pi}{2} \to 0$.

$(2) \,\,\,$ If $x-\dfrac{\pi}{2} \to 0$ then $-\Big(\dfrac{\pi}{2}-x\Big) \to 0$.

In this step, we have taken that $y = \dfrac{\pi}{2}-x$. Therefore, $-y \to 0$.

$(3) \,\,\,$ If $-y \to 0$ then $-y \times (-1) \to 0 \times (-1)$. Therefore, $y \to 0$.

Therefore, it is cleared that if $x$ approaches $\dfrac{\pi}{2}$, then $y$ approaches $0$. Now, write the whole given problem in terms of $y$ by eliminating the $x$.

$\implies$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{\cos{x}}{\dfrac{\pi}{2}-x}}$ $\,=\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{2}-y\Big)}}{y}}$

Simplify the Algebraic trigonometric function

The angle $\dfrac{\pi}{2}-y$ represents a variable angle, which belongs to the first quadrant. In the first quadrant, $\cos{\Big(\dfrac{\pi}{2}-y\Big)} = \sin{y}$.

Therefore, the function in the numerator $\cos{\Big(\dfrac{\pi}{2}-y\Big)}$ can be replaced by its equivalent value $\sin{y}$.

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

Evaluate the Limit of the function

According to the limit rules of trigonometric functions, the limit of sinx/x is equal to one as x approaches 0. Therefore, the limit of the $\dfrac{\sin{y}}{y}$ as $y$ tends to $0$ is also equal to $1$.

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

$= \,\,\,$ $1$

Therefore, it is evaluated that the limit of the ratio of $\cos{x}$ to $\dfrac{\pi}{2}-x$ is equal to $1$ as $x$ approaches $\dfrac{\pi}{2}$.



Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more