Question

Limits question:

Answer 1

\[1/\sqrt{x} \]

Answer 2

point (1,1)

Answer 3

Limit as x approaches 1? o.o

Answer 4

yup.

Answer 5

Direct substitution works fine here. You even gave your own answer. Since there exists a point at that x-value, that'll also be the limit of the function at x=1.

Answer 6

But, the answer is suppose to be -1/2

Answer 7

No it's not. :3
http://www.wolframalpha.com/input/?i=limit+as+x+approaches+1+of+1%2Fsqrt%28x%29

Answer 8

HMMMMMMMMMM. alright, thanks. lol

Answer 9

Where'd you get -1/2 from? lol

Answer 10

From my textbook. The answers at the back.

Answer 11

The lies are real. Interesting. Are you sure you wrote it right, or you're looking at the wrong question/answer?

Answer 12

I'm looking at the right question.

Answer 13

...The wrong answer? :3
Sorry, I'm not sure at all y dey lie.

Answer 14

Lol, alright, thanks.

Answer 15

Actualllllllllllly, we're suppose to use this equation: f(x)-f(a)/x-a. Then we're suppose to sub. in everything. and I get up until here: \[-\sqrt{x}/\sqrt{x}(x-1)\]
andddddd then you're suppose to sub. in 1 to see if it is one of the odd cases, which it is. But I don't know where to go from there.

Answer 16

what are you using for "a" ?

Answer 17

a would be 1.

Answer 18

and then f(a) would be 1 too :P

Answer 19

oh shoot i used the wrong f(x) :{

Answer 20

oops. :P

Answer 21

ok you should have
\[\frac{1-\sqrt{x}}{\sqrt{x}(x-1)}*\frac{(1+\sqrt{x})}{(1+\sqrt{x})} = \frac{-(x-1)}{\sqrt{x}(x-1)(1+\sqrt{x})}\]

Answer 22

yeah I did. now.

Answer 23

do*

Answer 24

what would my restriction be for \[1+\sqrt{x}\]

Answer 25

no restriction, you are taking limit at x=1

Answer 26

But, i always have to write what the denominator cannot be so it doesn't equal 0.

Answer 27

oh i see well "1 + sqrtx" is never 0 so its not an issue here

Answer 28

Alright.

Answer 29

Merci beaucoup!