Full solution

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

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We want to find

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

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What happens when we use direct substitution?

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Choose

1

answer:

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

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

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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 $x = -\frac{\pi}{4}$ .