Question

Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method,

Answer 1

\[(\sqrt{1+3x}-\sqrt{1+7x})/x\]

Answer 2

did you try rationalizing the numerator ?

Answer 3

so taking the derivative of the top and bottom i get
\[(1/(2\sqrt{1-3x}))-1/2\sqrt{1+7x}\]

Answer 4

no i didnt
i figure thatd be more difficult

Answer 5

multiply numerator and denominator by
\(\Large \sqrt{1+3x}+\sqrt{1+7x}\)

Answer 6

ok

Answer 7

or , once you took the derivative you can simply plug in x=0

Answer 8

but your derivative does not seem to be correct

Answer 9

one sec

Answer 10

but then i still have 0 on the bottom

Answer 11

you do ? how ??

Answer 12

well it says with hospitals law you can take the derivative of the numerator and then the derivative of the top......i do. because if x approaches 0, x(√1+3x+√1+7x) equals zero

Answer 13

You don't apply the quotient rule in L'H.
You find the derivative of the numerator separately and the derivative of the denominator separately.

Answer 14

L'Hopital's rrule say that you take the derivative of numerator and denominator separately.

the denominator is x
its derivative is =1

Answer 15

ah i see how my derivative could be wrong