Math / Calculus

Question6. Evaluate $\lim _{x \rightarrow \frac{\pi}{2}}(\sin x-\cos x)^{\tan x}$ [5 pts]

Studdy Solution

STEP 1

1. We are given the expression $(\sin x - \cos x)^{\tan x}$ .
2. We need to evaluate the limit as $x$ approaches $\frac{\pi}{2}$ .
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 $\sin x - \cos x$ and $\tan x$ as $x$ approaches $\frac{\pi}{2}$ .
2. Determine if the expression $(\sin x - \cos x)^{\tan x}$ 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 $\sin x - \cos x$ as $x \to \frac{\pi}{2}$ :
$\sin\left(\frac{\pi}{2}\right) = 1, \quad \cos\left(\frac{\pi}{2}\right) = 0$
Thus, $\sin x - \cos x \to 1 - 0 = 1$ .

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

Apply L'Hôpital's Rule:
Differentiate the numerator and the denominator:
Numerator: $\frac{d}{dx}[\ln(\sin x - \cos x)] = \frac{\cos x + \sin x}{\sin x - \cos x}$
Denominator: $\frac{d}{dx}[\cot x] = -\csc^2 x$
Apply L'Hôpital's Rule:
$\lim_{x \to \frac{\pi}{2}} \frac{\ln(\sin x - \cos x)}{\cot x} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x + \sin x}{(\sin x - \cos x)(-\csc^2 x)}$

STEP 9

Simplify the expression:
As $x \to \frac{\pi}{2}$ , $\sin x \to 1$ and $\cos x \to 0$ , so the expression simplifies to:
$\lim_{x \to \frac{\pi}{2}} \frac{\cos x + \sin x}{(\sin x - \cos x)(-\csc^2 x)} = \lim_{x \to \frac{\pi}{2}} \frac{1}{-1} = 0$

STEP 10

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

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