Question

as t approaches zero

(1/t * sqrt 1+t)-(1/t)

Answer 1

\[\frac{ 1 }{ t \sqrt{1+t} }-\frac{ 1 }{ t }\]?

Answer 2

yes!

Answer 3

multiply the 2nd(top and bottom) term by sqrt(1+t)

we have

(1-sqrt(1-t))/(t*sqrt(t-1))

ok?

Answer 4

ok

Answer 5

then devide every term t

((1/t)-(sqrt(1+t)/t)) /((t/t)sqrt(t+1)) =
((1/t)-sqrt(1+t)/t)/sqrt(t+1)
ok?

Answer 6

ps those sqrt(t-1) should be sqrt(t+1)

Answer 7

ye i was just about to ask why you used (t-1)

Answer 8

so whats the final answer?

Answer 9

you will stall have 2t on the bottom

Answer 10

take it twice:)

Answer 11

\[\frac{ 1 }{ t \sqrt{1+t} }-\frac{1}{ t }\]
add the fractions to get
\[ \frac{1-\sqrt{1+t}}{t\sqrt{1+t}} \]
as t ->0 this fraction approaches 0/0
so we can use L'Hopitals Rule
First, clear the sort by multiplying top and bottom by sqrt(1+t)
\[ \frac{\sqrt{1+t}-1-t}{t^2+t} \]
now take the derivative of top and bottom
and let t-->0

Answer 12

have you learned l`hopital's Rule?

Answer 13

@steel11 ?

Answer 14

nope

Answer 15

hmm

Answer 16

the bottom give 2t+1 --> 1 as t-->0

the top gives (1/2)(1+t)^(-1/2) -1 -->1/2-1 = -1/2

Answer 17

yeah, i made a mistake, but he has not learned L'Hapital's law yet.

Answer 18

rule*

Answer 19

thanks both of you for the help. the answer is -.5.

Answer 20

I would be curious to see how to do it without l'hopital.

Answer 21

yeah no idea

Answer 22

(1-sqrt(x+1))/(x*sqrt(x+1))

top and bottom by (1+sqrt(x+1))

1-(x+1)/(x*sqrt(x+1)+x(x+1))

devide every term by x

-1/(sqrt(x+1)+x+1)

run limit

=-1/2

@phi @steel11

Answer 23

3rd line should be (1-(x+1)/(x*sqrt(x+1)+x(x+1))

Answer 24

looks good