Question

hospitals rule for finding limits of indeterminate
1 mark for review
lim_{x\to\frac{\pi}{2}}\frac{3\cos x}{2x - \pi} is
a -\frac{3}{2}
b 0
c \frac{3}{2}
d nonexistent

Explanation:

Step1: Check the form of the limit

When \(x = \frac{\pi}{2}\), \(\cos(\frac{\pi}{2})=0\) and \(2x-\pi=2\times\frac{\pi}{2}-\pi = 0\). So, it is in the \(\frac{0}{0}\) indeterminate form.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and the denominator. The derivative of \(y = 3\cos x\) is \(y^\prime=-3\sin x\) (using the formula \((\cos x)^\prime=-\sin x\)), and the derivative of \(y = 2x-\pi\) is \(y^\prime = 2\) (using the formula \((ax + b)^\prime=a\)).

Step3: Evaluate the new limit

\(\lim_{x
ightarrow\frac{\pi}{2}}\frac{3\cos x}{2x - \pi}=\lim_{x
ightarrow\frac{\pi}{2}}\frac{-3\sin x}{2}\)
Substitute \(x=\frac{\pi}{2}\) into \(\frac{-3\sin x}{2}\), since \(\sin(\frac{\pi}{2}) = 1\), we get \(\frac{-3\times1}{2}=-\frac{3}{2}\)

Answer:

A. \(-\frac{3}{2}\)