lim sqrt(1-cosx)/|tanx| as x approaches 0

is there anyway of solving this without using lhopitals rule?
that was the first thing i did actually
Then wait instead of making a new thread. Btw, I'm pretty sure you have to do L'Hopital, but I'm too lazy to check. Buck up and get it over with, you're wasting time when you already know the solution.
our teacher assigned it, and we still haven't learned l'hopitals rule so there must be a wayy
I'd help you, but we're only on antiderivatives, and I haven't learned Ihopitals rule yet..

Try changing tan to sinx/cosx and look for trig properties.
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Last edited by Fat Lard at Oct 23, 2008,
Well, when in doubt guess 0, 1, or positive or negative infinity. 95% of limit questions end up being one of those four.
is sqrt(1-cosx)/|tanx| the function? so, ( f(x) = sqrt(1-cosx)/|tanx| )?

And, is just 1-cos(x) inside the square root, or is the whole rest of the thing?
f(x) = sqrt(1-cosx)/|tanx| is the function yes, only 1-cos(x) is in the square root
ok well they are dif coming from the left and the right. so the limit itself doesn't exist. or do you need the limits from the left and from the right?
ok wait actually i didn't remember the absolute value signs... i'm not sure if that changes it... we're working on limits right now in ap calc, but i have never heard of the l'hopital rule
I can never remember how to solve limits.
no man the limit exists and when i punch it in my graphing calc the answer is sqrt(2)/2
ok nm the limit is around .70711
yea I know the answer, just not how to get there
oh, so yeah square root of 2 over 2 makes sense because that's like .707106. But yeah I put it into my calculator and did the table to look at the x's as close to 0 as I could without getting them. But you can only look at the right side, since you can't put in absolute value into a calc. (At least, I can't...) If you could, then all of the |tanx| values would be positive, and that makes all the y-value results positive too. So if you do graph it you see that once you cross x=0, the y's become negative, but you just have to pretend like they're positive. So yeah the limit is (square root os 2)/2. Hope that made some sort of sense
Or, if you can't use a calculator then just test it with x-values close to 0. Like plug in -1, -.1, -.01, -.001, .001, .01, .1, and 1, and see what number it gets closest to as the number gets closer to 0.
Pi, the answer is always pi.
do you have a midterm tomorrow?
Quote by beatles33164
ok wait actually i didn't remember the absolute value signs... i'm not sure if that changes it... we're working on limits right now in ap calc, but i have never heard of the l'hopital rule

absolute value is the distance from a number to 0. so -7= 7 i and i have no clue for i'hopital sorry im in 9th grade honors algebra not calculus
if i could solve that, i think i would be able to rule the world...
Quote by IwantaTele
absolute value is the distance from a number to 0. so -7= 7 i and i have no clue for i'hopital sorry im in 9th grade honors algebra not calculus

Oh no I meant that I didn't remember that they were there when I put it into my calc. But I got it afterwards
sqrt(1-cosx)/|tanx| = sqrt(1-cosx)/sqrt((tanx)^2)
= sqrt((1-cosx)/((tanx)^2))
dropping the square root signs everywhere now - too tedious to type them in. Just know all these expressions are all being square rooted

= (1-cosx)(cosx)^2/(sinx)^2
= (1-cosx)(cosx)^2/(1-cosx^2)
= (1-cosx)(cosx)^2/(1-cosx)(1+cosx)
= (cosx)^2/(1+cosx)
= 1/sqrt(2)

edit: hope that makes any sense
thanks Jakinov, that helped me out