# L'hopital's rule

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?

VietDao29
Homework Helper
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 $$\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)}$$ is in one of the 2 Indeterminate Forms $$\frac{0}{0} ; \ \mbox{and } \frac{\infty}{\infty}$$, note that you can only use L'Hospital for (0/0), and (inf/inf).

If the requirement above meets, then:
$$\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}$$. 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:

$$\lim_{x \rightarrow 0} \left( \frac{1}{\sin (2x)} - \frac{1}{2x} \right) = \lim_{x \rightarrow 0} \left( \frac{2x - \sin (2x)}{2x \sin (2x)} \right)$$

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? :)

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Extra practice:

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

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. $$\lim_{x \rightarrow 0 ^ +} \ln (x)$$
b. $$\lim_{x \rightarrow 0 ^ +} x ^ {-5} = \lim_{x \rightarrow 0 ^ +} \frac{1}{x ^ 5}$$?

Can you go from here? :)

Last edited:
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.)

VietDao29
Homework Helper
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:
$$\lim_{x \rightarrow 0} \frac{\ln x}{x ^ 5}$$ is a little bit meaningless, it's not defined for non-positive x, so I think the problem probably ask for:
$$\lim_{x \rightarrow 0 ^ +} \frac{\ln x}{x ^ 5}$$

2a is correct.
2b is actually + infty
$$\lim_{x \rightarrow 0 ^ +} \frac{1}{x ^ 5} = + \infty$$
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?

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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 . I should go to bed now.

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