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Find
lim_(x rarr(pi)/(4))(cos(2x))/(cos(x)-sin(x))
Choose 1 answer:
(A)
sqrt2
(B) 2
(c) 4
(D) The limit doesn't exist

Find $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos (2 x)}{\cos (x)-\sin (x)}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\sqrt{2}$ $\newline$ (B) $2$ $\newline$ (C) $4$ $\newline$ (D) The limit doesn't exist

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Is (x, y) a solution to the system of equations?

Full solution

Q. Find

\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos (2 x)}{\cos (x)-\sin (x)}

.

\newline

Choose

1

answer:

\newline

(A)

\sqrt{2}

\newline

(B)

2

\newline

(C)

4

\newline

(D) The limit doesn't exist

Substitute Values: We need to find the limit of the function $(\cos(2x))/(\cos(x)-\sin(x))$ as $x$ approaches $\pi/4$ . Let's first try to directly substitute $x = \pi/4$ into the function to see if it yields a determinate form. $\newline$ $\cos(2 \times (\pi/4)) = \cos(\pi/2) = 0$ $\newline$ $\cos(\pi/4) = \sqrt{2}/2$ $\newline$ $\sin(\pi/4) = \sqrt{2}/2$ $\newline$ Now, substitute these values into the function: $\newline$ $(\cos(2 \times (\pi/4)))/(\cos(\pi/4) - \sin(\pi/4)) = 0/(\sqrt{2}/2 - \sqrt{2}/2) = 0/0$ $\newline$ We get an indeterminate form $0/0$ , which means we need to apply L'Hôpital's Rule or algebraic manipulation to find the limit.
Apply L'Hôpital's Rule: Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit of $f(x)/g(x)$ as $x$ approaches a value $c$ is $0/0$ or $\infty/\infty$ , then the limit of $f(x)/g(x)$ as $x$ approaches $c$ is the same as the limit of $f'(x)/g'(x)$ as $x$ approaches $c$ , provided that the latter limit exists. $\newline$ Let's find the derivatives of the numerator and the denominator. $\newline$ The derivative of $x$ $1$ with respect to $x$ is $x$ $3$ . $\newline$ The derivative of $x$ $4$ with respect to $x$ is $x$ $6$ .
Find Derivatives: Now we will apply L'Hôpital's Rule by taking the limit of the derivatives: $\newline$ $\lim_{x \to \frac{\pi}{4}}\left(\frac{-2\sin(2x)}{-\sin(x) - \cos(x)}\right)$ $\newline$ Let's substitute $x = \frac{\pi}{4}$ into the derivatives: $\newline$ $-2\sin(2 \times (\frac{\pi}{4})) = -2\sin(\frac{\pi}{2}) = -2$ $\newline$ $-\sin(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}$ $\newline$ Now, we substitute these values into the limit of the derivatives: $\newline$ $\frac{-2}{-\sqrt{2}} = \frac{2}{\sqrt{2}} = 2 \times (\frac{\sqrt{2}}{2}) = \sqrt{2}$
Apply L'Hôpital's Rule Again: We have found that the limit of the original function as $x$ approaches $\frac{\pi}{4}$ is $\sqrt{2}$ . Therefore, the correct answer is (A) $\sqrt{2}$ .

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