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Yakvenalex [24]
2 years ago
11

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim x → (π/2)+

cos(x) 1 − sin(x)
Mathematics
1 answer:
Otrada [13]2 years ago
5 0

Answer:

The limit of the function at x approaches to \frac{\pi}{2} is -\infty.

Step-by-step explanation:

Consider the information:

\lim_{x \to \frac{\pi}{2}}\frac{cos(x)}{1-sin(x)}

If we try to find the value at \frac{\pi}{2} we will obtained a \frac{0}{0} form. this means that L'Hôpital's rule applies.

To apply the rule, take the derivative of the numerator:

\frac{d}{dx}cos(x)=-sin(x)

Now, take the derivative of the denominator:

\frac{d}{dx}1-sin(x)=-cos(x)

Therefore,

\lim_{x \to \frac{\pi}{2}}\frac{-sin(x)}{-cos(x)}

\lim_{x \to \frac{\pi}{2}}\frac{sin(x)}{cos(x)}

\lim_{x \to \frac{\pi}{2}}tan(x)}

Since, tangent function approaches -∞ as x approaches to \frac{\pi}{2}

, therefore, the original expression does the same thing.

Hence, the limit of the function at x approaches to \frac{\pi}{2} is -\infty.

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