Expert: Paul Klarreich - 7/13/2007

Question
The question is lim x->0negative (tan x) / x^2
(oh and the limit at 0 with the little negative sign up the top i don't know how to write it here).

So i applied l'Hospital's rule to it and got

lim x-> 0neg (sec x)^2  /  2x

and that parts all good as now its not indeterminate.

What i don't understand is how to get the limit here. Do I use the graph of sec x? or... (the answer book says its  infinity. I got the infinity part via sketching but not sure how the sign is decided)

thanks.

Answer
Questioner:   kita
Category:  Calculus
Question:  The question is lim x->0negative (tan x) / x^2
(oh and the limit at 0 with the little negative sign up the top i don't know how to write it here).

So I applied l'Hospital's rule to it and got

lim x-> 0neg (sec x)^2  /  2x

and that part's all good as now it's not indeterminate.

What I don't understand is how to get the limit here. Do I use the graph of sec x? or... (the answer book says its  infinity. I got the infinity part via sketching but not sure how the sign is decided)

thanks.
.........................................
Hi, Kita,

For your example,

        tan x
lim    --------
x->0-     x^2

yes, l'Hospital's rule leads to:

      sec^2(x)
lim    --------
x->0-     2x

And you are correct -- this is no longer an indeterminate form, because:

sec(x) is continuous at  x = 0, since  cos(0) = 1, so sec(0) = 1, and also lim sec(x) = 1, by continuity.  So you have:

      sec^2(x)        1
lim    -------- -->  -----
x->0-     2x         2(0-)

Note that I wrote  0- in the denominator, because that's how you are taking the limit.  So this says you are taking very small NEGATIVE values of x.  In that case the fraction will be a very LARGE negative number, so you write  -infinity as the limit.