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PROBLEM

6. Evaluate limxπ2(sinxcosx)tanx\lim _{x \rightarrow \frac{\pi}{2}}(\sin x-\cos x)^{\tan x} [5 pts]

STEP 1

1. We are given the expression (sinxcosx)tanx(\sin x - \cos x)^{\tan x}.
2. We need to evaluate the limit as xx approaches π2\frac{\pi}{2}.
3. Direct substitution might lead to an indeterminate form, so we may need to use L'Hôpital's Rule or other limit properties.

STEP 2

1. Analyze the behavior of sinxcosx\sin x - \cos x and tanx\tan x as xx approaches π2\frac{\pi}{2}.
2. Determine if the expression (sinxcosx)tanx(\sin x - \cos x)^{\tan x} results in an indeterminate form.
3. Use logarithmic properties to simplify the expression if necessary.
4. Apply L'Hôpital's Rule or other limit techniques to evaluate the limit.
5. Simplify and find the final value of the limit.

STEP 3

Evaluate sinxcosx\sin x - \cos x as xπ2x \to \frac{\pi}{2}:
sin(π2)=1,cos(π2)=0 \sin\left(\frac{\pi}{2}\right) = 1, \quad \cos\left(\frac{\pi}{2}\right) = 0 Thus, sinxcosx10=1\sin x - \cos x \to 1 - 0 = 1.

STEP 4

Evaluate tanx\tan x as xπ2x \to \frac{\pi}{2}:
tanx=sinxcosx \tan x = \frac{\sin x}{\cos x} As xπ2x \to \frac{\pi}{2}, cosx0\cos x \to 0, so tanx\tan x \to \infty.

STEP 5

Check if (sinxcosx)tanx(\sin x - \cos x)^{\tan x} results in an indeterminate form:
Since (sinxcosx)1(\sin x - \cos x) \to 1 and tanx\tan x \to \infty, the expression 11^\infty is an indeterminate form.

STEP 6

Use logarithmic properties to simplify the expression:
Let y=(sinxcosx)tanx y = (\sin x - \cos x)^{\tan x} .
Take the natural logarithm:
lny=tanxln(sinxcosx) \ln y = \tan x \cdot \ln(\sin x - \cos x)

STEP 7

Evaluate the limit of lny\ln y as xπ2x \to \frac{\pi}{2}:
limxπ2lny=limxπ2tanxln(sinxcosx) \lim_{x \to \frac{\pi}{2}} \ln y = \lim_{x \to \frac{\pi}{2}} \tan x \cdot \ln(\sin x - \cos x) Since tanx\tan x \to \infty and ln(sinxcosx)ln(1)=0\ln(\sin x - \cos x) \to \ln(1) = 0, we have the form 0\infty \cdot 0.
Rewrite as a fraction:
limxπ2ln(sinxcosx)cotx \lim_{x \to \frac{\pi}{2}} \frac{\ln(\sin x - \cos x)}{\cot x}

STEP 8

Apply L'Hôpital's Rule:
Differentiate the numerator and the denominator:
Numerator: ddx[ln(sinxcosx)]=cosx+sinxsinxcosx\frac{d}{dx}[\ln(\sin x - \cos x)] = \frac{\cos x + \sin x}{\sin x - \cos x}
Denominator: ddx[cotx]=csc2x\frac{d}{dx}[\cot x] = -\csc^2 x
Apply L'Hôpital's Rule:
limxπ2ln(sinxcosx)cotx=limxπ2cosx+sinx(sinxcosx)(csc2x) \lim_{x \to \frac{\pi}{2}} \frac{\ln(\sin x - \cos x)}{\cot x} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x + \sin x}{(\sin x - \cos x)(-\csc^2 x)}

STEP 9

Simplify the expression:
As xπ2x \to \frac{\pi}{2}, sinx1\sin x \to 1 and cosx0\cos x \to 0, so the expression simplifies to:
limxπ2cosx+sinx(sinxcosx)(csc2x)=limxπ211=0 \lim_{x \to \frac{\pi}{2}} \frac{\cos x + \sin x}{(\sin x - \cos x)(-\csc^2 x)} = \lim_{x \to \frac{\pi}{2}} \frac{1}{-1} = 0

SOLUTION

Since limxπ2lny=0\lim_{x \to \frac{\pi}{2}} \ln y = 0, we have:
limxπ2y=e0=1 \lim_{x \to \frac{\pi}{2}} y = e^0 = 1 The value of the limit is:
1 \boxed{1}

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