Given that the domain is all real numbers, what is the limit of the range for the function ƒ(x) = 4^2x

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

How can you tell if 78 is divisible by 2 3 OR 6

noname [10]

If its an even number its divisible by 2 if you add the answer (7+8) and it equals a number (7+8=15) divisible by 3 then 78 is divisible by 3 and if it is divisible by both 2 and 3 then you know it is divisible by 6 but you have to be able to divide the number by BOTH 2 AND 3

4 0

3 years ago

76,361 deliveries/hour = deliveries/week

IrinaK [193]

Answer:

12,828,648 deliveries per week

Step-by-step explanation:

There are 24 hours in one day, and 7 days in a week. Multiply 24 with 7:

24 x 7 = 168

Next, multiply 76361 with 168:

76361 x 168 = 12828648

There are 12,828,648 deliveries per week.

~

3 0

3 years ago

ato must write 60 percent of a 20-page report by tomorrow. Kato wants to determine the number of pages that he needs to write by

ivanzaharov [21]

Answer:

Step 4

Step-by-step explanation:

because 100/20=5

so 60/5=12

8 0

3 years ago

Are 2,3,4,5,6,9,10 divisble by 3

babunello [35]

No, only 3, 6, and 9 are divisible by 3 because 3 x 1= 3, 3 x 2= 6, 3 x 3= 9. Hope this helps

8 0

3 years ago

Read 2 more answers

Given that the domain is all real numbers, what is the limit of the range for the function ƒ(x) = 4^2x - 100?