lim 1+(1/√x)/cot x
x→0
Hello,
I am working on my calc 1 homework where we use L'Hôpital's rule to evaluate limits. When I try to evaluate this problem to find the indeterminate form, it becomes undefined so I know I need to rewrite it in a different way, I just don't know how. All help is appreciated!
lim 1+(1/√x)
x→0 —————
cot x
2 answers
So, you have (1 + 1/√x)/cot x
As x→0, that goes to ∞/∞
So, if we try derivatives, we always get the same dang ∞/∞
So, let's rewrite things as
tanx (1+√x) / √x
Now as x→0, that goes to 0/0
Trying derivatives now, we get
[sec^2x(1+√x) + tanx(2/√x)] / (2/√x)
= sec^2x(1+√x)/(2/√x) + tanx*(2/√x) / (2/√x)
= [√x sec^2x(1+√x)]/2 + tanx
That goes to 0
As x→0, that goes to ∞/∞
So, if we try derivatives, we always get the same dang ∞/∞
So, let's rewrite things as
tanx (1+√x) / √x
Now as x→0, that goes to 0/0
Trying derivatives now, we get
[sec^2x(1+√x) + tanx(2/√x)] / (2/√x)
= sec^2x(1+√x)/(2/√x) + tanx*(2/√x) / (2/√x)
= [√x sec^2x(1+√x)]/2 + tanx
That goes to 0
6 answers