Mathematics, 07.07.2019 12:30 bloodgif3386

Could someone show me how to evaluate $\lim_{x \to \ 4+} \frac{\sqrt{x^{2}-16 } }{x-4}$ ?

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Answer from: lm18618

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.)

Answer from: Quest

Arhombus is a quadrilateral.

Answer from: Quest

54 kilometers per hour

step-by-step explanation:

according to rules:

1 km = 1000 meter

which means that 1 meter = 1/1000 km

1 hour = 3600 second [ as 1 hour = 60 min = 60 *60 seconds ]

which means that 1 second = 1/3600 hour

therefore, 1 kilometer/hour = $(1/1000)/(1/3600)=3600/1000=18/5=3.6 kilometer/hour$

thus to convert 15 meter per second to kilometer per hour, we multiply 15 to 3.6 kilometer/hour.

15 meter/second= $15*3.6=54 kilometer/hour$ which is the required answer.

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Could someone show me how to evaluate [tex]\lim_{x \to \ 4+} \frac{\sqrt{x^{2}-16 } }{x-4}[/tex]?...

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