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
lim 1+(1/√x)/cot x
x→0
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
To rewrite the given expression and apply L'Hôpital's Rule, we can start by simplifying the expression and then differentiate the numerator and denominator separately.
Let's break it down step by step:
1. Begin by simplifying the expression as much as possible:
lim (x → 0) [(1 + (1/√x))/cot(x)]
2. The function cot(x) can be expressed as cos(x)/sin(x).
lim (x → 0) [(1 + (1/√x))/cos(x)/sin(x)]
3. Next, simplify the expression by multiplying the numerator and denominator by sin(x):
lim (x → 0) [(1 + (1/√x))(sin(x))/cos(x)]
= lim (x → 0) [(sin(x) + (sin(x)/√x))/cos(x)]
4. Now, differentiate the numerator and denominator separately:
For the numerator, we differentiate using the product rule: d(uv) = u * dv + v * du.
d/dx (sin(x) + (sin(x)/√x)) = cos(x) + [cos(x)/√x] - (1/2)x^(-1/2)sin(x)
For the denominator, the derivative of cos(x) is -sin(x).
5. Rewrite the expression with the differentiated numerator and denominator:
lim (x → 0) [(cos(x) + [cos(x)/√x] - (1/2)x^(-1/2)sin(x)) / -sin(x)]
6. Now, we can evaluate the limit by directly substituting x = 0 into the expression:
[(cos(0) + [cos(0)/√0] - (1/2)0^(-1/2)sin(0)) / -sin(0)]
= [(1 + [1/0] - (1/2) * (0/0)) / 0]
Notice that the expression becomes undefined at this point.
7. Since we obtained an undefined form, we can rewrite the expression in a different way to apply L'Hôpital's Rule again.
Taking the derivative of the numerator and denominator separately, we have:
lim (x → 0) [(-sin(x) - (1/2)(-1/2)x^(-3/2)sin(x) - (1/2)x^(-1/2)cos(x))/ -cos(x)]
8. By substituting x = 0 into the expression, we have:
[(-sin(0) - (1/2)(-1/2) * 0^(-3/2)sin(0) - (1/2) * 0^(-1/2)cos(0))/ -cos(0)]
= [-sin(0) - (1/2)(-1/2) * (0/0)sin(0) - (1/2) * (0/0)cos(0))/ -cos(0)]
= [0/0]
We see that we once again obtained an undefined form, but this time in the form of 0/0.
9. To continue, we need to apply L'Hôpital's Rule again. Take the derivative of the numerator and denominator to get:
lim (x → 0) -cos(x)
10. Finally, substitute x = 0 into the expression:
-cos(0)
= -1
Therefore, the limit of the given expression as x approaches 0 is -1.