I was studying for my Math53 finals (eep) tomorrow when I came upon a function that has a non-existent limit (positive infinity from the right, negative infinity from the left: a vertical asymptote).
And then Mean Girls comes into mind. So, nerdy-ness kicks in and I rushed to Dad’s laptop to search for the limit-less function from Mean Girls fame.
So, this is what I found:
| lim x-> 0 |
ln(1-x) - sin(x) 1-cos2(x) |
Upon initial inspection of this function, one would notice that finding its limit is not an easy 30-second task. Substitution would yield an indeterminate limit, with a form of 0/0. So, how did the Mean Girl get it in less than a minute? Hoooow?
So I began to solve for the limit. What I first did was be to use l’Hospital’s rule, because the initial limit was in an indeterminate form. It would then yield
| lim x-> 0 |
-1/(1-x)-cos(x) 2cos(x)sin(x) |
But when we substitute 0 for x, we would get -2/0. So we need to check the left and right hand limits for us to be able to conclude if the limit exists (as either positive or negative ∞) or if it does not exist.
So…
| lim x-> 0+ |
-1/(1-x)-cos(x) 2cos(x)sin(x) |
| lim x-> 0- |
-1/(1-x)-cos(x) 2cos(x)sin(x) |
The above limit would yield a limit of +∞ while the limit below gives us a limit of -∞.
So, after a lot of computations (and much time), we can finally conclude that the limit does not exist since the left hand and right hand limits are not the same.
Madaya lang talaga si Cady kaya niya nasolve yun. Di kaya ng powers ko yun!
Oh my gulay. I need a life. Pray for my finals tomorrow.
UPDATE: Hindi ko pala kinailangang mag l’Hospital’s. Hahaha. Kaya palang gawin yun sa simpleng pagsubstitute ng maliliit na values from the left and the right of 0.


March 28th, 2007 at 9:49 am
BA! Haha, nag-aaral ako for math 100 tas lagi ko naaalala yung “the limit does not exist!” hehehe!
March 28th, 2007 at 6:37 pm
Wahahaha! Ang kulet noh. XD
March 29th, 2007 at 7:13 am
BA, is that you? what are you talking about? it’s all greek, or rather GEEK, to me!
March 29th, 2007 at 12:41 pm
Mama: May kulto ako. Ganyan mga kalokohan namin tuwing gabi. Ang kultong yun ay tinatawag naming CalCULTus. Wahahaha
April 15th, 2009 at 3:56 am
you can also see it just by considering the first terms of the expansions of each function!
top -x-x
bottom xsquared (since it is the square of sinx)
-2/x ‘limit does not exist’