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## Differential Calculus

### Course: Differential Calculus>Unit 4

Lesson 8: L’Hôpital’s rule: composite exponential functions

# L’Hôpital’s rule (composite exponential functions)

Sal uses L'Hôpital's rule to find the limit at 0 of (sinx)^(1/lnx).

## Want to join the conversation?

• When I got to the step that Sal does at I applied L'Hôpital's rule to the whole thing again. I got the same answer, but Sal pointed out a slightly easier way by using the rule that "the limit of the product of two functions is equal to the product of their limits."

I had forgotten that rule - is there a name for it, and a proof of it? • At Sal starts explaining how y should intuitively approach e, w/the expression lim x->0+ ln(y)=1. Right before that he shows that ln(y) = 1. Couldn't you instead just raise e to both sides of the equation and get e^ln(y) = e^1 thus y = e ? • 10.23, how is he taking the derivative without doing the quotient rule. is that a mistake? • No. He is not taking the derivative of the whole function, he is using l'Hopital's rule.
This rule involves (but only valid if the limit is of a 0/0 or ∞/∞ form) taking the derivative of the numerator divided by the derivative of the denominator NOT the derivative of the entire function.

In fact, with l'Hopital's rule, if you take the derivative of the whole function, you will get the wrong answer.

In summary, l'Hopitals rule states that IF you have
lim x→c f(x)/g(x)
AND the derivative of g(x) exists and is not 0.
AND one of the following is true:
f(x) and g(x) both approach 0
or
f(x) and g(x) both approach ±∞
THEN and only then:
lim x→c f(x)/g(x) = lim x→c f'(x)/g'(x)
This is NOT the derivative f(x) / g(x)
• At, , how is ln (sin x) ^ 1/ln (x) = {1/ln(x)} * ln (sin x) ? • At , Why couldn't we factorize `0^0` as `0^1 / 0^1`, which is 0/0 indeterminate value, since we know that `0^1/0^1 is 0^1-1` which is `0^0`. He says justifications, what does he mean? • at Sal suggested taking the natural log of both sides. Where are more videos on this and questions on natural log (ln) derivatives? • At around in the video, would it be valid to, instead of splitting up the single limit into two and taking the first derivative of one, just take the second derivative of x*cos(x) / sin(x) to get
cos(x) + (-x*sin(x)) / cos (x) ? When evaluating it, you still get 1 + 0 / 1, but is that just a coincidence? • If I followed what you meant correctly, yes, that is a valid way of doing it.
You can continue using multiple iterations of l'Hopital's provided the derivatives exist AND you continue to have 0/0 or ∞/∞ forms. If you have those forms, you do not necessarily have to split up the limit -- though doing so is often easier.

Remember that l'Hopital's is NOT valid or true if you don't have 0/0 or ∞/∞ . So, you must be careful about that.
• So, is this alright? Rather than stop at taking the natural log, I then raised everything to a power of e, so as to ignore the whole "y=" side of the equation. This resulted in:
e^((lnsinx)/lnx)
Which accepts an x-value of 0 in order to evaluate the limit being equal to e. • The `e^(ln(something))` is ultimately equivalent to `ln|y| = ln|something|`, because when you go to cancel the `ln|y| = ln|something|` you will use e: `e^(ln|y|) = e^(ln|something|)`

For your posted example, where you declare that `e^((ln|sin(x)|)/ln|x|)` when evaluated at 0 directly results in e, is incorrect. As illustrated in the @ , you have a `-∞/-∞` condition. You still need to cycle through the L'Hopital's iterations to the final resolvable state:

``L = lim  e^( (cos(x) - x*sin(x))/cos(x) )    x→0+L = e^( (cos(0) - 0*sin(0))/cos(0) )L = e^( 1 - 0*0)/1 )L = e^( 1/1 )L = e``  