Question

if (f(x)=\frac{sin x - 1}{cos^{2}x}), then (lim_{x
ightarrow\frac{pi}{2}}f(x)) is equivalent to which of the following?

Explanation:

Step1: Substitute $x = \frac{\pi}{3}$ into $f(x)$

First, rewrite $f(x)=\frac{\sin x - 1}{\cos^{2}x}$. When we try to directly substitute $x = \frac{\pi}{3}$, we get $\frac{\sin\frac{\pi}{3}-1}{\cos^{2}\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}-1}{(\frac{1}{2})^{2}}=\frac{\frac{\sqrt{3}-2}{2}}{\frac{1}{4}} = 2(\sqrt{3}-2)$. But we can also use the trig - identity $\cos^{2}x=1 - \sin^{2}x=(1 + \sin x)(1 - \sin x)$.

Step2: Rewrite the function

Rewrite $f(x)=\frac{\sin x - 1}{\cos^{2}x}=\frac{\sin x - 1}{1-\sin^{2}x}=\frac{\sin x - 1}{(1 + \sin x)(1 - \sin x)}=-\frac{1}{1+\sin x}$ (for $\sin x
eq1$).

Step3: Calculate the limit

Now, find $\lim_{x
ightarrow\frac{\pi}{3}}f(x)=\lim_{x
ightarrow\frac{\pi}{3}}-\frac{1}{1+\sin x}$. Substitute $x = \frac{\pi}{3}$ into $-\frac{1}{1+\sin x}$, we have $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, so $-\frac{1}{1+\frac{\sqrt{3}}{2}}=-\frac{2}{2 + \sqrt{3}}$. Rationalize the denominator: $-\frac{2(2-\sqrt{3})}{(2+\sqrt{3})(2 - \sqrt{3})}=-(4 - 2\sqrt{3})=2\sqrt{3}-4$.

Answer:

$2\sqrt{3}-4$