True or False: if the product of three numbers is negative, then all the numbers are negative?

Evaluate the limit with either L'Hôpital's rule or previously learned methods.lim Sin(x)- Tan(x)/ x^3x → 0

Vsevolod [243]

Answer:

$\dfrac{-1}{6}$

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3}$ , to evaluate the following steps must be carried out.

Step 1: substitute x = 0 into the function

$= \dfrac{sin(0)-tan(0)}{0^3}\\= \frac{0}{0} (indeterminate)$

Step 2: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the function

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ sin(x)-tan(x)]}{\frac{d}{dx} (x^3)}\\= \lim_{ x\to \ 0} \dfrac{cos(x)-sec^2(x)}{3x^2}\\$

Step 3: substitute x = 0 into the resulting function

$= \dfrac{cos(0)-sec^2(0)}{3(0)^2}\\= \frac{1-1}{0}\\= \frac{0}{0} (ind)$

Step 4: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 2

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ cos(x)-sec^2(x)]}{\frac{d}{dx} (3x^2)}\\= \lim_{ x\to \ 0} \dfrac{-sin(x)-2sec^2(x)tan(x)}{6x}\\$

$= \dfrac{-sin(0)-2sec^2(0)tan(0)}{6(0)}\\= \frac{0}{0} (ind)$

Step 6: Apply L'Hôpital's rule, by differentiating the numerator and denominator of the resulting function in step 4

$= \lim_{ x\to \ 0} \dfrac{\frac{d}{dx}[ -sin(x)-2sec^2(x)tan(x)]}{\frac{d}{dx} (6x)}\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^2(x)sec^2(x)+2sec^2(x)tan(x)tan(x)]}{6}\\\\= \lim_{ x\to \ 0} \dfrac{[ -cos(x)-2(sec^4(x)+2sec^2(x)tan^2(x)]}{6}\\$

Step 7: substitute x = 0 into the resulting function in step 6

$= \dfrac{[ -cos(0)-2(sec^4(0)+2sec^2(0)tan^2(0)]}{6}\\\\= \dfrac{-1-2(0)}{6} \\= \dfrac{-1}{6}$

Hence the limit of the function $\lim_{ x\to \ 0} \dfrac{sin(x)-tan(x)}{x^3} \ is \ \dfrac{-1}{6}$ .

3 0

3 years ago

I just need the boxes filled up i need help

nirvana33 [79]

Answer:

8 is how much you get paid ( sorry if I'm wrong)

8 0

3 years ago

Mike thought of a number. He added to it to two other numbers. One of these numbers was 12 less than his number, the other was 1

Inessa [10]

Answer:

number 1 + number 2 + number 3 = 15

Number 1 = x + 12

Number 2 = x - 12

Number 3 = x

(x + 12) + (x - 12) + x = 15

x + 12 + x - 12 + x = 15

x + x + x +12 -12 = 15

3x = 15

x = 15/3 = 5

Check:

Number 1 = 17

Number 2 = 5 -12 = -7

Number = 5

number 1 + number 2 + number 3 = 15

17 + (-7) + 5 = 15

10 + 5 = 15

15 = 15

Answer is 5

Step-by-step explanation:

5 0

3 years ago

a company that manufactures bicycles has a fixed cost of $80,000. it costs $100 to produce each bicycle. the total cost for the

frez [133]

C(x) = 80000 + 100x is the total cost as function of number of cycles produced

C(90) = 89000 and it costs $ 89000 to produce 90 bicycles

Solution:

Given that, company that manufactures bicycles has a fixed cost of $80,000

Fixed cost = $ 80,000

Let x be the number of cycles produced

Let C(x) be the total cost as function of number of cycles produced

It costs $100 to produce each bicycle

Variable cost = 100 x number of cycles produced

variable cost = 100x

The total cost for the company is the sum of its fixed cost and variable costs

total cost = fixed cost + variable cost

C(x) = 80000 + 100x

Thus total cost as function of "x" is found

Find and interpret C(90)

Substitute x = 90 in C(x)

C(90) = 80000 + 100(90)

C(90) = 80000 + 9000

C(90) = 89000

Thus it costs $ 89000 to produce 90 bicycles

3 0

3 years ago

Identify the fraction that is equivalent to 3/8 A. 15/32 B. 12/32 C. 12/24 D. 9/32

Burka [1]

Answer:

The answer is B

Step-by-step explanation:

plz mark brainliest i answered first

4 0

2 years ago

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