Question

lim x^2/×-2 ×-->2

Answer 1

i think the answer is 2x as x --> 2 is 4

Answer 2

you should use the l'hopital's rule

Answer 3

the what rule¿

Answer 4

I graphed it and got two curves sort of circling (0, 0) intersecting at-2 and 2

Answer 5

Are u starting calc? have do done derivatives yet

Answer 6

my assignment is due today, these lim questions are the last questions and I havent covered lim yet so im battling through blindly hehe

Answer 7

that is clear, but how do I know to use that rule? what's the give away?

Answer 8

Isn't your limit
\[\lim_{x\to2}\frac{x^2}{x-2x}?\]

Answer 9

yes

Answer 10

Never mind, you can't use L'Hopital's Rule...

Answer 11

In that case, L'hopital's rule (whether or not you've learned it) cannot be applied. Direct substitution should work, though.

Answer 12

no!!!! just x-2

Answer 13

\[\lim_{x\to2}\frac{x^2}{x-2}\]

Answer 14

yes

Answer 15

In that case, I think this is one of those problems you have to work on systematically. Meaning, figure out the value of the function when \(x=1.9, 1.99, 1.999,\ldots,\) and when \(x=2.1, 2.01,2.001,\ldots\) .

In other words, you examine the one-sided limits,
\[\lim_{x\to2^-}\frac{x^2}{x-2}\text{ and }\lim_{x\to2^+}\frac{x^2}{x-2}\]
systematically. I'm not sure there are any algebraic manipulations/tricks you can use.

Answer 16

I did that, and gtaphed it.

Answer 17

graph is curve (4, 8) (1,-1) through x at 2
curve (-3,-8) (0, 2) through x at -2

Answer 18

Okay. I don't have the luxury of graphing quickly at the moment, so I'll try to lead you through this one with my own reasoning. As x approaches 2 from the left, the numerator will always be positive, but the denominator will be negative (since, for instance, 1.9 - 2 = -0.1). The denominator will become smaller (but still negative) as x gets closer and closer to 2, so the fraction itself will approach negative infinity.

As x approaches 2 from the right, the denominator will be positive (2.000...1 - 2 = 0.00...1 > 0). (Numerator is still positive.) The denominator gets smaller and smaller, so the fraction approaches positive infinity. So, since the limits don't match, the limit as x approaches 2 does not exist.

Answer 19

and des my graph support that?

Answer 20

I think so, yeah. You have a vertical asymptote at x = 2, with the function curving downward on the left and upward on the right (relative to the asymptote).

Answer 21

you can use google to plot it. type into the search window
plot x^2/(x-2)

but you should be able to sketch it by hand.

Answer 22

I did sketch it phi . see avoce comment

Answer 23

err above

Answer 24

yes, but you said
I graphed it and got two curves sort of circling (0, 0) intersecting at-2 and 2

which does not sound correct. You can check your work and see if you did it correctly.
the graph should show the curves shooting up to infinity on one side of 2 and going down to minus infinity on the other side of 2

Answer 25

ok I'll double check my calculations.