Expert: Paul Klarreich - 9/13/2009

Question
QUESTION: im just asking for clarification
if you have lim x approaching infinity (x.tan(1/x))
and you l hospital getting
sin(1/x)/cos(1/x).(1/x)
does the sin(1/x)/(1/x) become 1 or zero?
1, because lim x approaching infinity sin(1/x)/cos(1/x).(1/x)
is the same as lim t approaching 0+ sin(t)/t which is the formula
or
zero, due to squeeze theorem. (which is confusing me because
-1/x < sinx/x < 1/x doesn't necessarily mean sinx/x=0.. or does it) D:<

ANSWER: Questioner: Aaron
Country: United States
Category: Calculus
Private: No
Subject: Calculus BC limits
Question: im just asking for clarification

if you have lim [x -> infinity] (x * tan(1/x))
and you l hospital getting

>> I guess anything can be a verb these days.

sin(1/x)/cos(1/x).(1/x)

does the sin(1/x)/(1/x) become 1 or zero?
1, because lim x approaching infinity sin(1/x)/cos(1/x).(1/x)
is the same as lim [t -> 0+] sin(t)/t = 1 which is the formula

>> I think this is the right one.

or

0, due to squeeze theorem. (which is confusing me because
-1/x < sinx/x < 1/x doesn't necessarily mean sinx/x=0.. or does it)

This does not look relevant, but I think your idea of letting t = 1/x is a good one that you could have applied earlier.

Try this:

If  t = 1/x, then  x->inf means t -> 0.

lim [x -> inf] (x * tan(1/x)) becomes

lim [t -> 0] (tan(t)/t) =
         sin t   1
lim [t -> 0] ----- ----- -->
         t   cos t
     1
1 * ----- = 0, and you don't even have to 'lope'
     1
(which you are not supposed to know in September)

---------- FOLLOW-UP ----------

QUESTION: at the end 1*1=0?
ehh...
or it was just a typo? :D

ooh and the thing about me using t=1/x.
doesnt zero have to be approached from the right (so its 0+) which isnt a formula.. that's what i was asking ahaha :]
or that isnt relevant. :O

Answer
Of course it is.  
         1
Indeed,  1 * - = 1
         1

Sorry about that.

And, yes,   x -> +infinity  would mean  t -> 0+
But   lim[x -> 0] sin x / x = 1 (well-known limit), so

lim[x -> 0+] sin x / x = 1
and
lim[x -> 0-] sin x / x = 1