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QuestionFind lim⁡x→π4g(x)\lim _{x \rightarrow \frac{\pi}{4}} g(x) for g(x)=cos⁡x−sin⁡x1−2sin⁡2xg(x)=\frac{\cos x-\sin x}{1-2 \sin ^{2} x}. Options: (A) 0 (B) 12\frac{1}{\sqrt{2}} (C) 2\sqrt{2} (D) Limit does not exist.

Studdy Solution

STEP 1

Assumptions1. The function g(x)g(x) is defined as g(x)=cos⁡x−sin⁡x1−sin⁡xg(x)=\frac{\cos x-\sin x}{1- \sin ^{} x} . We are asked to find the limit of g(x)g(x) as xx approaches π4\frac{\pi}{4}

STEP 2

First, we substitute x=π4x=\frac{\pi}{4} into the function g(x)g(x).
g(π4)=cos⁡π4−sin⁡π41−2sin⁡2π4g\left(\frac{\pi}{4}\right)=\frac{\cos \frac{\pi}{4}-\sin \frac{\pi}{4}}{1-2 \sin ^{2} \frac{\pi}{4}}

STEP 3

We know that cos⁡π=sin⁡π=12\cos \frac{\pi}{} = \sin \frac{\pi}{} = \frac{1}{\sqrt{2}}. Substitute these values into the equation.
g(π)=12−121−2(12)2g\left(\frac{\pi}{}\right)=\frac{\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}}{1-2 \left(\frac{1}{\sqrt{2}}\right)^{2}}

STEP 4

implify the numerator and the denominator separately.
g(π4)=01−2×12g\left(\frac{\pi}{4}\right)=\frac{0}{1-2 \times \frac{1}{2}}

STEP 5

Further simplify the denominator.
g(π4)=01−1g\left(\frac{\pi}{4}\right)=\frac{0}{1-1}

STEP 6

implify the fraction.
g(π4)=00g\left(\frac{\pi}{4}\right)=\frac{0}{0}This is an indeterminate form, so we need to use L'Hopital's rule to find the limit.

STEP 7

Apply L'Hopital's rule, which states that if the limit of a function is in the form 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}, then the limit of that function is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.
lim⁡x→π4g(x)=lim⁡x→π4g′(x)h′(x)\lim{x \rightarrow \frac{\pi}{4}} g(x) = \lim{x \rightarrow \frac{\pi}{4}} \frac{g'(x)}{h'(x)}where g′(x)g'(x) is the derivative of the numerator and h′(x)h'(x) is the derivative of the denominator.

STEP 8

Calculate the derivative of the numerator and the denominator.
g′(x)=−sin⁡x−cos⁡xg'(x) = -\sin x - \cos xh′(x)=−4sin⁡xcos⁡xh'(x) = -4 \sin x \cos x

STEP 9

Substitute the derivatives into the limit.
lim⁡x→π4−sin⁡x−cos⁡x−4sin⁡xcos⁡x\lim{x \rightarrow \frac{\pi}{4}} \frac{-\sin x - \cos x}{-4 \sin x \cos x}

STEP 10

Substitute x=π4x=\frac{\pi}{4} into the limit.
lim⁡x→π4−sin⁡π4−cos⁡π4−4sin⁡π4cos⁡π4\lim{x \rightarrow \frac{\pi}{4}} \frac{-\sin \frac{\pi}{4} - \cos \frac{\pi}{4}}{-4 \sin \frac{\pi}{4} \cos \frac{\pi}{4}}

STEP 11

Substitute sin⁡π4=cos⁡π4=\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{}{\sqrt{}} into the limit.
lim⁡x→π4−−−4××\lim{x \rightarrow \frac{\pi}{4}} \frac{-\frac{}{\sqrt{}} - \frac{}{\sqrt{}}}{-4 \times \frac{}{\sqrt{}} \times \frac{}{\sqrt{}}}

STEP 12

implify the limit.
lim⁡x→π4−2−2=2\lim{x \rightarrow \frac{\pi}{4}} \frac{-\sqrt{2}}{-2} = \sqrt{2}So, the limit of g(x)g(x) as xx approaches π4\frac{\pi}{4} is 2\sqrt{2}.

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