YACT (Yet another Calc Thread)

[SXMP]

A fairly simple one, that is driving me crazy:

lim xsec(x)csc(x)
x->0

I know the answer is 1, can anyone figure out how to get there?

BTW I already tried googling it to no avail.

 

[RaynorWolfcastle]

x/[sin(x)cos(x)]
0/0 form, apply L'hopital's rule to get
1/[cos(x)^2 - sin(x)^2]
which is 1/1

done

 

[SXMP]

Hmm, yeah L'hopital's rule does work. I didn't notice it though because *supposedly* we are supposed to be able to solve this problem with the sinx/x=1 identity.

Can you think of anyway to solve it that way. (BTW I've already taken Calc, I'm just retaking it in college and want to be able to do the problems on the quiz the way they showed us so far... because *technically* I'm not supposed to know L'hopital's rule

 

[aux]

Originally posted by: SXMP
Hmm, yeah L'hopital's rule does work. I didn't notice it though because *supposedly* we are supposed to be able to solve this problem with the sinx/x=1 identity.

Can you think of anyway to solve it that way. (BTW I've already taken Calc, I'm just retaking it in college and want to be able to do the problems on the quiz the way they showed us so far... because *technically* I'm not supposed to know L'hopital's rule

sin(x)/x != 1
sin(x)/x -> 1 when x-> 0

about your problem: knowing that sin(x)/x -> 1 when x->0 is enough because cos(0) = 1, so there is no problem with the cos at 0 (i.e. no division by 0).

 

[Excelsior]

Check this link out for your future questions concerning Calc.

 

[RaynorWolfcastle]

Originally posted by: aux

Originally posted by: SXMP
Hmm, yeah L'hopital's rule does work. I didn't notice it though because *supposedly* we are supposed to be able to solve this problem with the sinx/x=1 identity.

Can you think of anyway to solve it that way. (BTW I've already taken Calc, I'm just retaking it in college and want to be able to do the problems on the quiz the way they showed us so far... because *technically* I'm not supposed to know L'hopital's rule

sin(x)/x != 1
sin(x)/x -> 1 when x-> 0

about your problem: knowing that sin(x)/x -> 1 when x->0 is enough because cos(0) = 1, so there is no problem with the cos at 0 (i.e. no division by 0).

yeah, all you do is say that the limit of products is the product of limits so that

lim x/sin(x)cos(x) = lim x/sin(x)*lim 1/cos(x)
= lim [sin(x)/x]^-1 * 1
= (lim[sin(x)/x])^-1
= 1^-1 = 1

 

[SithSolo1]

OMG, I hate Calc. Failing it and the lab right now. Been through 2 tutors, spent $300 on guide books and software, still nothing. I just can't do upper level math.