h(x)=(cot^(2)(x))/(1+sqrt2sin(x)) We want to find lim_(x rarr-(pi)/(4))h(x). What happens when we use direct substitution? Choose 1 answer: (A) The limit exists, and we found it! (B) The limit doesn't exist (probably an asymptote). (C) The result is indeterminate.

Direct Substitution: Let's first try direct substitution of x = -π/4 into the function h(x) to see what we get. h(x) = (cot^2(x)) / (1 + √2sin(x)) h(-π/4) = (cot^2(-π/4)) / (1 + √2sin(-π/4))Cotangent of -π/4: We know that cot(x) = 1/tan(x), and tan(-π/4) = -1. Therefore, cot(-π/4) = -1. cot^2(-π/4) = (-1)^2 = 1Denominator of h(x): Now let's look at the denominator of h(x) when x = -π/4. 1 + √2sin(-π/4) = 1 + √2(-1/√2) = 1 - 1 = 0Division by Zero: Since the denominator becomes 0, we have a division by zero situation, which means the limit does not exist due to a potential asymptote or discontinuity at x = -π/4.

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h(x)=(cot^(2)(x))/(1+sqrt2sin(x))
We want to find
lim_(x rarr-(pi)/(4))h(x).
What happens when we use direct substitution?
Choose 1 answer:
(A) The limit exists, and we found it!
(B) The limit doesn't exist (probably an asymptote).
(C) The result is indeterminate.

$h(x)=\frac{\cot ^{2}(x)}{1+\sqrt{2} \sin (x)}$ $\newline$ We want to find $\lim _{x \rightarrow-\frac{\pi}{4}} h(x)$ . $\newline$ What happens when we use direct substitution? $\newline$ Choose $1$ answer: $\newline$ (A) The limit exists, and we found it! $\newline$ (B) The limit doesn't exist (probably an asymptote). $\newline$ (C) The result is indeterminate.

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Full solution

h(x)=\frac{\cot ^{2}(x)}{1+\sqrt{2} \sin (x)}

\newline

We want to find

\lim _{x \rightarrow-\frac{\pi}{4}} h(x)

\newline

What happens when we use direct substitution?

\newline

Choose

1

answer:

\newline

(A) The limit exists, and we found it!

\newline

(B) The limit doesn't exist (probably an asymptote).

\newline

Direct Substitution: Let's first try direct substitution of $x = -\frac{\pi}{4}$ into the function $h(x)$ to see what we get. $\newline$ $h(x) = \frac{(\cot^2(x))}{(1 + \sqrt{2}\sin(x))}$ $\newline$ $h(-\frac{\pi}{4}) = \frac{(\cot^2(-\frac{\pi}{4}))}{(1 + \sqrt{2}\sin(-\frac{\pi}{4}))}$
Cotangent of $-\frac{\pi}{4}$ : We know that $\cot(x) = \frac{1}{\tan(x)}$ , and $\tan(-\frac{\pi}{4}) = -1$ . Therefore, $\cot(-\frac{\pi}{4}) = -1$ . $\newline$ $\cot^2(-\frac{\pi}{4}) = (-1)^2 = 1$
Denominator of h(x): Now let's look at the denominator of h(x) when $x = -\frac{\pi}{4}$ . $1 + \sqrt{2}\sin(-\frac{\pi}{4}) = 1 + \sqrt{2}(-\frac{1}{\sqrt{2}}) = 1 - 1 = 0$
Division by Zero: Since the denominator becomes $0$ , we have a division by zero situation, which means the limit does not exist due to a potential asymptote or discontinuity at $x = -\frac{\pi}{4}$ .