Question

find the following limit. is the function continuous at the point being approached?
\\(\lim_{x\to\pi/4}\sqrt{\csc^{2}x + 2\tan x}\\)

choose the correct answer below and fill in the answer box within your choice. (simplify your answer.)
\\(\circ a.\\) \\(\lim_{x\to\pi/4}\sqrt{\csc^{2}x + 2\tan x}=\\), and the function is not continuous at \\(x = \frac{\pi}{4}\\).
\\(\circ b.\\) \\(\lim_{x\to\pi/4}\sqrt{\csc^{2}x + 2\tan x}=\\), and the function is continuous at \\(x = \frac{\pi}{4}\\).
\\(\circ c.\\) the limit does not exist, so the function is not continuous at \\(x = \frac{\pi}{4}\\).

Explanation:

Step1: Recall trigonometric - function values at $x = \frac{\pi}{4}$

We know that $\csc x=\frac{1}{\sin x}$ and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, so $\csc\frac{\pi}{4}=\sqrt{2}$, and $\tan\frac{\pi}{4} = 1$.

Step2: Substitute $x=\frac{\pi}{4}$ into the function

Substitute $x = \frac{\pi}{4}$ into $\sqrt{\csc^{2}x + 2\tan x}$. We get $\sqrt{(\sqrt{2})^{2}+2\times1}$.

Step3: Simplify the expression

First, $(\sqrt{2})^{2}=2$, then $2 + 2\times1=4$. So $\sqrt{(\sqrt{2})^{2}+2\times1}=\sqrt{4}=2$.

Step4: Check continuity

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a}f(x)=f(a)$. Since the function $y=\sqrt{\csc^{2}x + 2\tan x}$ is well - defined at $x=\frac{\pi}{4}$ and $\lim_{x
ightarrow\frac{\pi}{4}}\sqrt{\csc^{2}x + 2\tan x}=2$, and $f(\frac{\pi}{4}) = 2$, the function is continuous at $x=\frac{\pi}{4}$.

Answer:

B. $\lim_{x
ightarrow\pi/4}\sqrt{\csc^{2}x + 2\tan x}=2$, and the function is continuous at $x=\frac{\pi}{4}$.