Answer:
The limit of the function at x approaches to [tex]\frac{\pi}{2}[/tex] is [tex]-\infty[/tex].
Step-by-step explanation:
Consider the information:
[tex]\lim_{x \to \frac{\pi}{2}}\frac{cos(x)}{1-sin(x)}[/tex]
If we try to find the value at [tex]\frac{\pi}{2}[/tex] we will obtained a [tex]\frac{0}{0}[/tex] form. this means that L'Hôpital's rule applies.
To apply the rule, take the derivative of the numerator:
[tex]\frac{d}{dx}cos(x)=-sin(x)[/tex]
Now, take the derivative of the denominator:
[tex]\frac{d}{dx}1-sin(x)=-cos(x)[/tex]
Therefore,
[tex]\lim_{x \to \frac{\pi}{2}}\frac{-sin(x)}{-cos(x)}[/tex]
[tex]\lim_{x \to \frac{\pi}{2}}\frac{sin(x)}{cos(x)}[/tex]
[tex]\lim_{x \to \frac{\pi}{2}}tan(x)}[/tex]
Since, tangent function approaches -∞ as x approaches to [tex]\frac{\pi}{2}[/tex]
, therefore, the original expression does the same thing.
Hence, the limit of the function at x approaches to [tex]\frac{\pi}{2}[/tex] is [tex]-\infty[/tex].
