Question

limit exists or not? : Question.

Answer 1

Which limit?

Answer 2

\[\lim_{X \rightarrow -\infty} \frac{ 3x-8 }{ x^3+15x-9 }\]

\[\lim_{x \rightarrow \infty} \frac{ x^2-2x-11 }{ 13-3x^2 }\]

\[\infty \lim_{x \rightarrow -\infty}\frac{ Sqrt(x^2+14x) }{ }\]

Answer 3

Sqrt(x^2+14)/14-13x

Answer 4

find ' horizontal asymptote

Answer 5

@Luis_Rivera could you help me

Answer 6

For the first and second one, divide the numerator and denominator by the highest power of x, and use the fact the limit of a quotient it the quotient of it's limits.

For the last one, I'm not sure what it's supposed to be, is it
\[\lim_{x \rightarrow \infty} \frac{\sqrt{x^2 + 14x}}{14-13x}\]?

Answer 7

yes thats right

Answer 8

Ok so for the first one, you divide the numerator and denominator by the highest power of x, which is x^3
\[\Large \lim_{x \rightarrow -\infty} \frac{ 3x-8 }{ x^3+15x-9 }=\]
\[\Large \lim_{x \rightarrow -\infty} \frac{\frac{3}{x^2} - \frac{8}{x^3}}{1 + \frac{15}{x^2} - \frac{9}{x^3}}=\]
\[\Large \frac{\lim_{x \rightarrow -\infty} \left(\frac{3}{x^2} - \frac{8}{x^3}\right)}{\lim_{x \rightarrow -\infty} \left( 1 + \frac{15}{x^2} - \frac{9}{x^3}\right)}= \frac{0}{1} = 0\]

You should be able to do the second one yourself using the same method

Answer 9

thats cool, thanks.
what about the first one and second

Answer 10

I did the first one in the reply, you should try to do the second one yourself though and if you get stuck just tell me :)

Answer 11

ok! i will try and if i stuck please there for me

Answer 12

mm Do I mutpily by x^2? beacuse i got 0 for the second one.

Answer 13

You divide the numerator and denominator by the highest power of x in the denominator, which is x^2

Answer 14

Ok, this is what I did

Answer 15

\[\frac{ \frac{ x ^2}{ x^2 }-\frac{ 2x }{ x^2 }-\frac{ 11 }{ x^2 } }{ \frac{ 13 }{ x^2 } -\frac{ 3x^2 }{ x^2 }}\]

Answer 16

Ok that is correct so far, but you can simplify it into
\[\frac{1-\frac{2}{x}-\frac{11}{x^2}}{\frac{13}{x^2} - 3}\]

Now use the fact that
\[\lim_{x \rightarrow \infty} \frac{1-\frac{2}{x}-\frac{11}{x^2}}{\frac{13}{x^2} - 3} =
\frac{\lim_{x \rightarrow \infty}\left(1-\frac{2}{x}-\frac{11}{x^2}\right)}{\lim_{x \rightarrow \infty} \left(\frac{13}{x^2} - 3 \right)}\]

to solve it just take the limits :)

Answer 17

I see. I didnt thought that x^2/X^2 is 1. so it should be -(1/3)

Answer 18

Exactly :)

Answer 19

for the third one, it has sqrt root so i cnt just multuily x^2?

Answer 20

no, multipley by sqrt(x^2-14x?)

Answer 21

Yes, you should bring 14 - 13x into the radical by writing it as
\[\lim_{x \rightarrow \infty} \frac{\sqrt{x^2 + 14x}}{14-13x} =
\lim_{x \rightarrow \infty} \sqrt{\frac{x^2 + 14x}{(-13x + 14)^2}}\]
And then use the power law:
\[\lim_{x \rightarrow \infty} \sqrt{\frac{x^2 + 14x}{(-13x + 14)^2}} =
\sqrt{\lim_{x \rightarrow \infty} \frac{x^2 + 14x}{(-13x + 14)^2}}\]

You should be able to solve it by expanding the brackets and taking the limits as you did in the second problem

Answer 22

expanding brackets by mutiply x^2?

Answer 23

(-13x + 14)^2 = 169x^2 - 364x +196

so it becomes
\[\sqrt{\lim_{x \rightarrow \infty} \frac{x^2 + 14x}{169x^2 - 364x + 196}}\]

Answer 24

so it should be -(1/169)

Answer 25

? hope its right :/

Answer 26

\[-\sqrt{\frac{1}{169}}\]

Answer 27

Thank yo a lot, I enjoyed solving problem with you

Answer 28

oh and one more question if the limit was -. is it still the same answer for the thirdone