Limit question - please help?

Question

hi!

i've been trying to solve this for a while and I'm stuck!

using l'hopital

lim (ln(cos(x^2)))/(x^4)

x->0

the problem reads: ln(cosx^2) divided by x^4

Answer 1

.....................Good Luck!

Answer 2

lim [ ln( cos(x^2) ) / (x^4) ]

x -> 0

Using L'Hospital's rule, since the form is ln(1)/0, or 0/0, we get

lim [ { ( 1/cos(x^2) )  ( -sin(x^2) ) (2x) } / { 4x^3 } ]

x -> 0

Which we can manipulate.

lim [ -2x sin(x^2)/cos(x^2) ] / [ 4x^3 ]

x -> 0

lim [ -2x tan(x^2) ] / [ 4x^3 ]

x -> 0

And we get a cancellation.

lim [ -2 tan(x^2) ] / [ 4x^2 ]

x -> 0

If we plug in x = 0, we get tan(0)/0, or 0/0, so we may again apply L'Hospital's rule.

lim ( [ -2 sec^2(x^2) (2x) ] / [ 8x ] )

x -> 0

lim ( [ -4x sec^2(x^2) ] / [ 8x ] )

x -> 0

lim ( [ -sec^2(x^2) ] / [ 2 ] )

x -> 0

And look, we can plug in x = 0 directly now.

[ -sec^2(0^2) ] / 2

[ -sec^2(0) ] / 2

[ -(sec(0))^2 ] / 2

[ -(1)^2 ] / 2

-1/2

Answer 3

lim(x->0) [-2x sin(x^2)/cos(x^2)] / 4x^3 (L'Hopital's Rule once)

= - lim(x->0) [tan(x^2)] / 2x^2

= - lim(x->0) [2x sec^2 (x^2)]/ 4x (L'Hopital's Rule again)

= (-1/2) lim(x->0) [sec^2 (x^2)]

= (-1/2)