evaluate the following limit. lim x→π/2 csc x - 1 / cot² x select the correct choice below and, if…

Explanation:

Step1: Rewrite in terms of sine and cosine

Recall that $\csc x=\frac{1}{\sin x}$ and $\cot x = \frac{\cos x}{\sin x}$. So the limit $\lim_{x\rightarrow\frac{\pi}{2}}\frac{\csc x - 1}{\cot^{2}x}=\lim_{x\rightarrow\frac{\pi}{2}}\frac{\frac{1}{\sin x}-1}{\frac{\cos^{2}x}{\sin^{2}x}}$. Simplify the complex - fraction: $\lim_{x\rightarrow\frac{\pi}{2}}\frac{\frac{1 - \sin x}{\sin x}}{\frac{\cos^{2}x}{\sin^{2}x}}=\lim_{x\rightarrow\frac{\pi}{2}}\frac{1 - \sin x}{\sin x}\cdot\frac{\sin^{2}x}{\cos^{2}x}=\lim_{x\rightarrow\frac{\pi}{2}}\frac{(1 - \sin x)\sin x}{1-\sin^{2}x}$.

Step2: Factor the denominator

Since $1-\sin^{2}x=(1 + \sin x)(1 - \sin x)$, the limit becomes $\lim_{x\rightarrow\frac{\pi}{2}}\frac{(1 - \sin x)\sin x}{(1 + \sin x)(1 - \sin x)}$. Cancel out the common factor $(1 - \sin x)$ (for $x\neq\frac{\pi}{2}$), we get $\lim_{x\rightarrow\frac{\pi}{2}}\frac{\sin x}{1+\sin x}$.

Step3: Substitute the value of $x$

Substitute $x = \frac{\pi}{2}$ into $\frac{\sin x}{1+\sin x}$. We know that $\sin\frac{\pi}{2}=1$. So $\frac{\sin\frac{\pi}{2}}{1+\sin\frac{\pi}{2}}=\frac{1}{1 + 1}=\frac{1}{2}$.

Answer:

A. $\frac{1}{2}$