L'Hospital's Rule
Lim(lnx)^(x-1)
x->1+
This is what I did so far...
Form: 0^0
ln(lnx)/ (1/x-1)
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I am not sure if you could use L'Hospital Rule (as stated in the instruction)for the 2 problems below:
lim [ln(y^2+2y)]/[lny]=1
y->0+
lim xlnx = 0
x->0+
I was able to solve them without using the Rule.
Actually, I would not recommend using any rules unless you really understand why and how they work. I.e. you should be able to derive those rules from first principles. Otherwise you are just training to become some sort of math computer who does not really understand math :)
L'Hopital's rule is actually closely related to Taylor's theorem which says that foir a sufficiently differentiable function f one has:
f(x+h) = f(x) + h f'(x) + h^2/2 f''(x) + h^3/6 f'''(x) + higher order terms in h.
This is something that you can easily derive and intuitively understand. Near a point x the function f will be f(x) plus correction terms. You can apply this to compute the limits as follows:
In case of (lnx)^(x-1) you take the logarithm, using that the log of the limit is the limit of the log which follows from the fact the the logarithm is a continuous function.
You then get the function
(x-1)ln(ln(x))
of which you want to compute the limit x--->1
You then put x = 1 + h and do a series expansion in powers of h. You get:
h(ln(ln(1+h)))
ln(1 + h) = h - h^2/2 + O(h^3)
where O(h^n) means a term proportional to h^n for small h. Taking the log again and using the same series expansion gives:
ln(ln(1 + h)) =
ln(h - h^2/2 + O(h^3)) =
ln(h) + ln(1 - h/2 +O(h^2)) =
ln(h) - h/2 + O(h^2)
Multiplying this by h and taking the limit h -->0 gives -1/2, so the original limit is exp(-1/2)
Note that we needed to go to second order in the Taylor expansion, which means that if you had used L'Hopital you would have had to use that rule twice.
Sorry, I was a bit confused, when multiplying by h the limit becomes zero so after exponentiationg you get 1. So, You could have used L'Hopital's rule once...
im going to figure this one out!
To answer your first question about using L'Hospital's Rule for the limit:
lim (ln(x)^(x-1))
x->1+
You are correct in rewriting it as 0^0. However, you made an error in the subsequent steps. Let's go through the correct steps together:
1. Start by taking the natural logarithm of the given expression:
ln[lim (ln(x)^(x-1))]
x->1+
2. Apply properties of logarithms to simplify further:
lim [(x - 1)ln(ln(x))]
x->1+
3. Now, the limit is in an indeterminate form 0 * infinity. This is where L'Hospital's Rule can be used.
Apply L'Hospital's Rule by differentiating the numerator and denominator:
lim [d/dx(x - 1)ln(ln(x))]/[d/dx(1)]
x->1+
4. Simplify the derivative:
lim [(ln(x) + (x - 1)/ln(x))(1/1)]
x->1+
5. Evaluate the limit:
[(ln(1) + (1 - 1)/ln(1))] / 1
= [0 + 0/0] / 1
= 0/1
= 0
So, the limit of ln(x)^(x-1) as x approaches 1 from the right side is 0.
Now, moving on to your second question, let's determine whether L'Hospital's Rule can be used for the limits:
1. lim [ln(y^2 + 2y)]/[ln(y)]
y->0+
In this case, you can simplify the expression without using L'Hospital's Rule:
ln[(y^2 + 2y)/(y)] = ln(y + 2)
Now, let's find the limit:
lim ln(y + 2)
y->0+
As y approaches 0 from the positive side, ln(y + 2) approaches ln(2), which is a finite value. Therefore, L'Hospital's Rule is not needed for this limit, and the answer is 1.
2. lim xln(x)
x->0+
Again, this limit can be solved without using L'Hospital's Rule:
Since ln(0) is undefined, we can rewrite the expression as:
lim ln(x)/(-1/x)
x->0+
Now, let's find the limit:
lim ln(x)/(-1/x)
x->0+
As x approaches 0 from the positive side, ln(x)/(-1/x) approaches ln(0)/(-∞) which is an indeterminate form. Therefore, L'Hospital's Rule can be applied here.
Differentiate the numerator and denominator:
lim [d/dx(ln(x))]/[d/dx(-1/x)]
x->0+
Simplify the derivatives:
lim [1/x]/[1/x^2]
x->0+
Evaluate the limit:
[1/0]/[1/0^2]
= ∞/∞
= undefined
So, using L'Hospital's Rule, the limit of xln(x) as x approaches 0 from the positive side is undefined.
In summary, you were able to correctly solve the limits without using L'Hospital's Rule. However, L'Hospital's Rule could have been applied to the second problem. It is always good to consider different approaches to solve limits, and L'Hospital's Rule is a helpful tool in certain cases of indeterminate forms.