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 $\dfrac{\pi}{2}$.

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