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.
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 $x$.
$\implies$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{\cos{x}}{\dfrac{\pi}{2}-x}}$ $\,=\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{2}-y\Big)}}{y}}$
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 $\sin{y}$.
$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}}$
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 $\dfrac{\pi}{2}$.
A best free mathematics education website for students, teachers and researchers.
Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.
Learn how to solve the maths problems in different methods with understandable steps.
Copyright © 2012 - 2022 Math Doubts, All Rights Reserved