A little calculus help...

[iamshady]

This might seem really easy for some of you, but can you give me a hand with these limits? Trying to take it apart to solve:

lim(cscx - cotx)
x->0

lim(1+ 3/x + 5/x^2)^x
x->infinity

Possible solution for the second one:

lim xln(1+ 3/x + 5/x^2) = (ln(x^2+3x+5))/(1/x) H (1/(x^2+3x+5))/(-1/x^2) = x^2/(x^2+3x+5) H 2x/(2x+3) H 2/2 = 1
x->infinity

H = Application of L'Hospital's Rule

Thanks in advance 🙂

 

[iamshady]

anyone?

 

[BruinEd03]

for the first one...try changing csc and cot in terms of sin and cos. THen try merging them into one fraction...then try l'hopitals.

-Ed

 

[jbahseng]

First question :

lim (cosecx- cotx) = lim (1/sinx - cosx/sinx)
x->0 x->inf
=lim (1-cosx)/sinx
x->inf
=lim sinx/cosx
x->inf
use L'Hopital's rule = 0/1
=0

 

[SnowPunk98]

/me + math = stupid ^1000

 

[BruinEd03]

<< First question :

lim (cosecx- cotx) = lim (1/sinx - cosx/sinx)
x->0 x->inf
=lim (1-cosx)/sinx
x->inf
=lim sinx/cosx
x->inf
use L'Hopital's rule = 0/1
=0 >>



NOt to burst ur bubble or anything...but u want x->0 not x->inf....sin and cos have no limits as x->inf (they keep oscillating) 😛

-Ed

 

[iamshady]

Thanks for the help with the first one guys 🙂

 

[jbahseng]

<<

<< First question :

lim (cosecx- cotx) = lim (1/sinx - cosx/sinx)
x->0 x->inf
=lim (1-cosx)/sinx
x->inf
=lim sinx/cosx
x->inf
use L'Hopital's rule = 0/1
=0 >>



NOt to burst ur bubble or anything...but u want x->0 not x->inf....sin and cos have no limits as x->inf (they keep oscillating) 😛

-Ed >>



Ouch! can i get partial credit? pleaseeeeeeeee😉

 

[SnowPunk98]

lol you guys are silly I hate math I would never take calc or trig or anything if I didnt have to for that matter

 

[BruinEd03]

<< Ouch! can i get partial credit? pleaseeeeeeeee😉 >>



Yes you can...I"m a softie hehe (I hate it when prof's don't give partial credit 🙁) 😀

-Ed

 

[DuffmanOhYeah]

<< /me + math = stupid ^1000 >>



LMAO! 🙂

 

[manuelku]

I hope that I won't ask you guys the same question when I take cal next sem 😎

 

[Bluga]

l'hopitals

 

[JayHu]

In case you havent solved these yet (and because i love easy math!)
i solved them for you:
1)
limx->0 (cscx-cotx)
= 1/sinx - 1/tanx
= 1/sinx - cosx/sinx
= (1-cosx)/sinx
= (1-cos^2x)/sinx * 1/(1+cosx) (multiplied by (1+cosx)/(1+cosx))
= sin^2x/sinx*1/1+cosx
= sinx * 1/1+cosx
= 0

2)
limx->inf (1+3/x+5/x^2)^x
= limx->inf e^(x log(1+3/x+5/x^2)) <- call this equation 1
im just going to look at the exponent part.
xlog( 1+3/x+5/x^2)
= x( log( 1/x^2(x^2+3x+5))
= x( log(1/x^2) + log(x^2+3x+5) )
= x( log(x^2+3x+5) - log(x^2) )
as x->inf
log(x^2+3x+5) - log(x^2) = 0
thus going back to equation 1
e^0 = 1 which is the answer..

any questions? i dont like using l'hopitals rule.. its the pansy way to solve limits! 😉

hopefully im right on this! 🙂