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.

Substituting x into h(x): Let's attempt to substitute x with -(pi)/(4) directly into the function h(x) to see what happens. h(x) = (cot^(2)(x))/(1+sqrt2sin(x)) h(-(pi)/(4)) = (cot^2(-(pi)/(4)))/(1+sqrt2sin(-(pi)/(4)))Evaluating cot^2(-(pi)/(4)): We need to evaluate cot^2(-(pi)/(4)) and sqrt2sin(-(pi)/(4)). The cotangent of -(pi)/(4) is the reciprocal of the tangent of -(pi)/(4). Since tan(-(pi)/(4)) = -1, cot(-(pi)/(4)) = -1.Evaluating sqrt2sin(-(pi)/(4)): Now we square cot(-(pi)/(4)) to get cot^2(-(pi)/(4)) = (-1)^2 = 1.Substituting values into h(x): Next, we evaluate sqrt2sin(-(pi)/(4)). Since sin(-(pi)/(4)) = -sin((pi)/(4)) and sin((pi)/(4)) = sqrt(2)/2, we have sqrt2sin(-(pi)/(4)) = sqrt(2) * (-sqrt(2)/2) = -1.Division by zero and the limit: Substitute these values back into the function h(x): h(-(pi)/(4)) = 1 / (1 - 1) This simplifies to h(-(pi)/(4)) = 1 / 0, which is undefined.Since we have a division by zero, the limit does not exist due to a vertical asymptote at x = -(pi)/(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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Math Problems

Algebra 1

Domain and range of relations

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

Substituting $x$ into $h(x)$ : Let's attempt to substitute $x$ with $-\frac{\pi}{4}$ directly into the function $h(x)$ to see what happens. $\newline$ $h(x) = \frac{\cot^{2}(x)}{1+\sqrt{2}\sin(x)}$ $\newline$ $h\left(-\frac{\pi}{4}\right) = \frac{\cot^2\left(-\frac{\pi}{4}\right)}{1+\sqrt{2}\sin\left(-\frac{\pi}{4}\right)}$
Evaluating $\cot^2\left(-\frac{\pi}{4}\right)$ : We need to evaluate $\cot^2\left(-\frac{\pi}{4}\right)$ and $\sqrt{2}\sin\left(-\frac{\pi}{4}\right)$ . The cotangent of $-\frac{\pi}{4}$ is the reciprocal of the tangent of $-\frac{\pi}{4}$ . Since $\tan\left(-\frac{\pi}{4}\right) = -1$ , $\cot\left(-\frac{\pi}{4}\right) = -1$ .
Evaluating $\sqrt{2}\sin\left(-\frac{\pi}{4}\right)$ : Now we square $\cot\left(-\frac{\pi}{4}\right)$ to get $\cot^2\left(-\frac{\pi}{4}\right) = (-1)^2 = 1$ .
Substituting values into $h(x)$ : Next, we evaluate $\sqrt{2}\sin\left(-\frac{\pi}{4}\right)$ . Since $\sin\left(-\frac{\pi}{4}\right) = -\sin\left(\frac{\pi}{4}\right)$ and $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ , we have $\sqrt{2}\sin\left(-\frac{\pi}{4}\right) = \sqrt{2} \times \left(-\frac{\sqrt{2}}{2}\right) = -1$ .
Division by zero and the limit: Substitute these values back into the function $h(x)$ : $h\left(-\frac{\pi}{4}\right) = \frac{1}{1 - 1}$ This simplifies to $h\left(-\frac{\pi}{4}\right) = \frac{1}{0}$ , which is undefined.
Division by zero and the limit: Substitute these values back into the function $h(x)$ : $h\left(-\frac{\pi}{4}\right) = \frac{1}{1 - 1}$ This simplifies to $h\left(-\frac{\pi}{4}\right) = \frac{1}{0}$ , which is undefined.Since we have a division by zero, the limit does not exist due to a vertical asymptote at $x = -\frac{\pi}{4}$ .