Mathematics, 10.03.2021 16:00 zariah10191101

What is this answer? Someone help me out with the answer please

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Answer from: Quest

n+3

step-by-step explanation:

we are increasing our answer by 3 every time. we are moving from negatives to positives, which can make it look confusing, but we are indeed increasing by 3. our next to numbers would be 4, 7, and 10.

Answer from: Quest

1.16 is the answer if i'm wrong u don't have to me

Answer from: Quest

"i dont know how to show the work, but the answer is 3 152/999 u have 1 decimal number repeating, u put it over u have 2 decimal numbers repeating, u put it over u have 3 decimal numbers repeating, u put it over dont really understand the work to get just know that is what u do "that is how it works !

Answer from: Quest

Answer

$\displaystyle\lim_{x\to 4^+} \dfrac{\sqrt{x^2-16}}{x-4} = \infty$

explanation

right now, attempting to use the quotient limit law is not possible since both the individual limits of the numerator and denominator are 0.

note that we can rationalize the numerator by multiplying the numerator and the denominator by $\sqrt{x^2-16}$ . our motivation is to get rid of the roots so that we can try to cancel (x-4) since that is what causes the 0/0 situation.

$\displaystyle\lim_{x\to 4^+} \dfrac{\sqrt{x^2-16}}{x-4} \cdot \dfrac{\sqrt{x^2-16}}{\sqrt{x^2-16}} \\ \\= \lim_{x\to 4^+} \dfrac{x^2-16}{(x-4)\sqrt{x^2-16}}$

since $x^2-16 = (x+4)(x-4)$

$\displaystyle= \lim_{x\to 4^+} \dfrac{(x+4)(x-4)}{(x-4)\sqrt{(x+4)(x-4)}} \\ \\= \lim_{x\to 4^+} \dfrac{x+4}{\sqrt{(x+4)(x-4)}}$

this cancellation $(x-4)/(x-4) = 1$ is only valid for x ≠ 4, but it is okay here since the limit only takes into account values around x = 4, not at x = 4.

now we make the following observation:

if x is close to 4 but more than 4, we have that

(x+4) approaches 4+4=8, but (x-4) is a small positive number, denoted by 0⁺

this ultimately means that

the numerator is approaching 8, and the denominator is approaching a small positive number, since in (x+4)(x-4) ≈ 8(0⁺) ≈ 0⁺. a small positive number times any number is still going to give you a small number. and then square rooting small number is still a small number.

so overall, we have 8/(small positive number), meaning that the entire fraction

$\dfrac{(x+4)}{\sqrt{(x+4)(x-4)}}$

is a large positive number. intuitively we get our answer:

$\displaystyle\lim_{x\to 4^+} \dfrac{\sqrt{x^2-16}}{x-4} = \infty$

(note: this does not mean that the limit exists, nor are we seeing ∞ as an actual number. this above is just shorthand for saying that the values of the expression grow arbitrarily large as x gets closer to a, but more than a.)

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