CALCU-Evaluate the limit, _x 1^+[(x-1) (2 x)].

Solution by Steps

step 1

We need to evaluate the limit

\lim _{x \rightarrow 1^{+}} \left[(x-1) \tan \left(\frac{\pi}{2} x\right)\right]

step 2

As

x

approaches

1

from the right,

x-1

approaches

0

from the right

step 3

We know that

\tan \left(\frac{\pi}{2} x\right)

approaches

\tan \left(\frac{\pi}{2}\right)

, which is undefined because

\tan \left(\frac{\pi}{2}\right)

tends to

+\infty

as

x

approaches

1

from the right

step 4

Therefore, we need to analyze the behavior of the product

(x-1) \tan \left(\frac{\pi}{2} x\right)

as

x

approaches

1

from the right

step 5

Let

t = x - 1

. As

x \rightarrow 1^{+}

,

t \rightarrow 0^{+}

. Thus, the limit becomes

\lim_{t \rightarrow 0^{+}} t \tan \left(\frac{\pi}{2} (t + 1)\right)

step 6

Simplify the argument of the tangent function:

\frac{\pi}{2} (t + 1) = \frac{\pi}{2} t + \frac{\pi}{2}

step 7

Therefore, the limit is

\lim_{t \rightarrow 0^{+}} t \tan \left(\frac{\pi}{2} t + \frac{\pi}{2}\right)

step 8

Using the trigonometric identity

\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

, we get

\tan \left(\frac{\pi}{2} t + \frac{\pi}{2}\right) = -\cot \left(\frac{\pi}{2} t\right)

step 9

As

t \rightarrow 0^{+}

,

\cot \left(\frac{\pi}{2} t\right) \rightarrow \frac{2}{\pi t}

step 10

Therefore,

\tan \left(\frac{\pi}{2} t + \frac{\pi}{2}\right) \approx -\frac{2}{\pi t}

step 11

Substituting this back into the limit, we get

\lim_{t \rightarrow 0^{+}} t \left(-\frac{2}{\pi t}\right) = \lim_{t \rightarrow 0^{+}} -\frac{2}{\pi} = -\frac{2}{\pi}

Answer

-\frac{2}{\pi}

Key Concept

Limit of a product involving a tangent function approaching an undefined value

Explanation

The key to solving this limit is recognizing the behavior of the tangent function as it approaches an undefined value and using trigonometric identities to simplify the expression.

Solution by Steps

step 1

To find the absolute extreme values of

f(x) = 3x(x+4)^{\frac{2}{3}}

on the interval

[-5, -1]

, we first need to find the critical points by taking the derivative of

f(x)

step 2

The derivative of

f(x)

is found using the product rule and chain rule:

f'(x) = 3(x+4)^{\frac{2}{3}} + 3x \cdot \frac{2}{3}(x+4)^{-\frac{1}{3}}

step 3

Simplifying the derivative:

f'(x) = 3(x+4)^{\frac{2}{3}} + 2x(x+4)^{-\frac{1}{3}}

step 4

Set the derivative equal to zero to find critical points:

3(x+4)^{\frac{2}{3}} + 2x(x+4)^{-\frac{1}{3}} = 0

step 5

Multiply through by

(x+4)^{\frac{1}{3}}

to clear the fraction:

3(x+4) + 2x = 0

step 6

Solve for

x

:

3x + 12 + 2x = 0 \implies 5x + 12 = 0 \implies x = -\frac{12}{5}

step 7

Evaluate

f(x)

at the critical point and the endpoints of the interval

[-5, -1]

:

step 8

f(-5) = 3(-5)(-1)^{\frac{2}{3}} = -15

step 9

f(-1) = 3(-1)(3)^{\frac{2}{3}} = -3 \cdot 3^{\frac{2}{3}}

step 10

f\left(-\frac{12}{5}\right) = 3\left(-\frac{12}{5}\right)\left(\frac{8}{5}\right)^{\frac{2}{3}}

step 11

Compare the values to determine the absolute maximum and minimum:

f(-5) = -15

,

f(-1) = -3 \cdot 3^{\frac{2}{3}}

, and

f\left(-\frac{12}{5}\right)

Answer

The absolute minimum value is

-15

at

x = -5

. The absolute maximum value is

-3 \cdot 3^{\frac{2}{3}}

at

x = -1

.

Key Concept

Finding absolute extreme values on a closed interval

Explanation

To find the absolute extreme values of a function on a closed interval, evaluate the function at critical points and endpoints, then compare the values.

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