L'Hopital's Rule: Solving Limits of x Approaching 0

In summary: I now get the infin. multiplication part now. also,f'(x)=ln(x)g'(x)=x^5 was meant to be:f(x) = ln(x)g(x) = x ^ 5And when you differentiate them, you get:f'(x) = 1/xg'(x) = 5x ^ 4So, the limit is:\lim_{x \rightarrow 0 ^ +} \frac{1/x}{5x ^ 4} = \lim_{x \rightarrow 0 ^ +} \frac{1}{5x ^ 5} = + \inftySo, there are 2 ways
  • #1
cd246
30
0
I would like to know if I did these the correct way.

Homework Statement


1.lim x->0 (1/sin 2x)-1/2x. the answer is 0.
2.lim x->0 (x^-5*ln x). The answer is -infinity.


Homework Equations


1.I used L'Hopitals theorem
2.I derived them, than L'hopitals.


The Attempt at a Solution


1.I only derived only the denominators for it to become (1/2cos2x)-(1/2), apply the 0 and it is (1/2)-(1/2)=0
2.f'(x) (-5x^-6)(1/x).
L'hopital's, f''(x) (30/x^7)or(30x^-7)
since the exponent is negative, is this the reason the answer is negative infinity?
 
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  • #2
cd246 said:
I would like to know if I did these the correct way.

Homework Statement


1.lim x->0 (1/sin 2x)-1/2x. the answer is 0.
2.lim x->0 (x^-5*ln x). The answer is -infinity.

Homework Equations


1.I used L'Hopitals theorem
2.I derived them, than L'hopitals.

The Attempt at a Solution


1.I only derived only the denominators for it to become (1/2cos2x)-(1/2), apply the 0 and it is (1/2)-(1/2)=0

Err... No, you cannot apply the rule like that. =.="

L'Hospital states that, if the limit [tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)}[/tex] is in one of the 2 Indeterminate Forms [tex]\frac{0}{0} ; \ \mbox{and } \frac{\infty}{\infty}[/tex], note that you can only use L'Hospital for (0/0), and (inf/inf).

If the requirement above meets, then:
[tex]\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}[/tex]. Note that, we differentiate both numerator, and denominator.

You cannot just differentiate the denominator, and leave the numerator unchanged. Well, that's not permitted. And you must have only one fraction, f(x)/g(x).

So, what you should do is to change it into a single fraction, like this:

[tex]\lim_{x \rightarrow 0} \left( \frac{1}{\sin (2x)} - \frac{1}{2x} \right) = \lim_{x \rightarrow 0} \left( \frac{2x - \sin (2x)}{2x \sin (2x)} \right)[/tex]

From here, there are 2 ways to do this, one is to use L'Hopital Rule, the other way is to use Taylor expansion for x around 0. They both work.

What Indeterminate Form is the expression above in? Is it 0/0, or inf/inf? Can you apply the rule now?

Can you go from here? :)

-------------------

Extra practice:

Find the limit of:
[tex]\lim_{x \rightarrow 0} \left( \frac{1}{\sin ^ 2 x} - \frac{1}{x ^ 2} \right)[/tex]

2.f'(x) (-5x^-6)(1/x).
L'hopital's, f''(x) (30/x^7)or(30x^-7)
since the exponent is negative, is this the reason the answer is negative infinity?

No, no, no, again, you should read the L'Hospital Rule again, L'Hopital Rule only defined for fraction, and not multiplication. You should make it a fraction first, before applying the rule.

Or, you can do it slightly different. Say:

1. Is the above expression valid for x <= 0?

2. What are the limits of:
a. [tex]\lim_{x \rightarrow 0 ^ +} \ln (x)[/tex]
b. [tex]\lim_{x \rightarrow 0 ^ +} x ^ {-5} = \lim_{x \rightarrow 0 ^ +} \frac{1}{x ^ 5}[/tex]?

Can you go from here? :)
 
Last edited:
  • #3
For 1. I would have to apply l'hopital's again because it is still 0/0. So, it would be(2-2cos(2x))/(4xcos2x+2sin2x). It is still 0/0 so i have to derive one more time. It would be 4sin2x/4xcos2x+4cos2x. Apply the 0 and got 0/4 = 0. Did I do this right now?

2. a. The limit of ln(x) is -infin
b. This is undefined.
So I would have to go on.
f'(x)=ln(x)
g'(x)=x^5

=1/5x^5. I use l'hopital and get -1/x^6 and I believe this will be infinate recursion. so the answer would be infin. or -infin.(as said in the book.)
 
  • #4
cd246 said:
For 1. I would have to apply l'hopital's again because it is still 0/0. So, it would be(2-2cos(2x))/(4xcos2x+2sin2x). It is still 0/0 so i have to derive one more time. It would be 4sin2x/4xcos2x+4cos2x. Apply the 0 and got 0/4 = 0. Did I do this right now?

Yup, this is good. :)
But your last step is wrong, you should re-check it. :)

2. a. The limit of ln(x) is -infin
b. This is undefined.

No, ln(x) is only define for x > 0, right? So what the problem asks for:
[tex]\lim_{x \rightarrow 0} \frac{\ln x}{x ^ 5}[/tex] is a little bit meaningless, it's not defined for non-positive x, so I think the problem probably ask for:
[tex]\lim_{x \rightarrow 0 ^ +} \frac{\ln x}{x ^ 5}[/tex]

2a is correct.
2b is actually + infty
[tex]\lim_{x \rightarrow 0 ^ +} \frac{1}{x ^ 5} = + \infty[/tex]
What is the result of 1/0.1, 1/0.001, 1/0.0000000000000000001?

So, what happens if you multiply the two? i.e: -infty, and +infty?

So I would have to go on.
f'(x)=ln(x)
g'(x)=x^5

=1/5x^5. I use l'hopital and get -1/x^6 and I believe this will be infinate recursion. so the answer would be infin. or -infin.(as said in the book.)

This is good, but what do you mean by f'(x), and g'(x)? You haven't differentiated them yet, so they should read f(x), and g(x) respectively.

And you are differentiating x5 in the denominator, how did u get -1/x6?

-------------------

Sorry, but I am a little bit rush, so it may be a little bit unclear. If you still wonder about anything, just shout it out, I or others may help. It's well past midnight, 3AM here :cry: :cry: :cry:. I should go to bed now. :cry: :cry: :cry: :cry:
 
  • #5
I now get the infin. multiplication part now.
also,
f'(x)=ln(x)
g'(x)=x^5 was meant to be:


f'(x)=ln(x)
-----------
g'(x)=x^5

Thanks
 

1. What is L'Hopital's Rule and when is it used?

L'Hopital's Rule is a mathematical theorem that allows us to solve limits of functions as the input (x) approaches a specific value (usually 0 or infinity). It is used when the limit of a function cannot be determined through direct substitution or other algebraic methods.

2. How does L'Hopital's Rule work?

L'Hopital's Rule states that if we have a limit of the form "f(x)/g(x)" where both f(x) and g(x) approach 0 (or infinity), then the limit of this expression is equal to the limit of the derivative of f(x) divided by the derivative of g(x). This process can be repeated multiple times until we reach a limit that can be evaluated directly.

3. What are the conditions for using L'Hopital's Rule?

The conditions for using L'Hopital's Rule are that both the numerator and denominator of the expression should approach 0 (or infinity) as the input (x) approaches the limit value. Additionally, the limit of the derivative of the numerator divided by the derivative of the denominator should exist and be finite.

4. Can L'Hopital's Rule be used for all limits?

No, L'Hopital's Rule can only be used for limits involving indeterminate forms such as 0/0 or infinity/infinity. If the limit is not an indeterminate form, then L'Hopital's Rule cannot be applied.

5. Are there any other methods for solving limits besides L'Hopital's Rule?

Yes, there are other methods for solving limits such as direct substitution, factoring, and rationalizing the numerator or denominator. However, L'Hopital's Rule is particularly useful for solving limits involving indeterminate forms and can often provide a quicker solution.

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