Can I use L'Hospitals rule here? (seem like i use it too often )

  • Thread starter linuxux
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In summary, the conversation is discussing the use of l'Hopital's rule to find the derivative of x^{4/3} at x=8. The conversation concludes that l'Hopital's rule cannot be used in this case, and instead suggests using the inverse function theorem. The conversation also touches on the differentiability of x^n and its derivative being equal to n x^(n-1).
  • #1
linuxux
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Does l'hospitals rule work here?:

[tex]\lim_{h\to_0}\frac{f\left(8+h\right)-f\left(8\right)}{h}[/tex] for [tex]f\left(x\right)=x^\frac{4}{3}[/tex]

then i would get,

[tex]\lim_{h\to_0}\frac{f\left(8+h\right)}{1}[/tex]

then,

[tex]=\frac{\left(8\right)^\frac{4}{3}}{1}[/tex] ?

is that it?...
 
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  • #2
No, you can't use it there, in my opinion. The question is asking you to work out the derivative of x^{4/3} at x=8, thus you can't invoke l'Hopital which a priori assumes the function to be differentiable. A minor quibble, but given the way you've written it it seems clear that you're being asked to prove the derivative exists, so assuming it does is not allowed.
 
  • #3
i tried factoring according to this:

[tex]a^4-b^4=\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)[/tex]

but I am stuck because i can't get a multiple of h to cancel the denomiator's h.
 
  • #4
Where did the power of 1/3 vanish to? (Note I don't have a simple solution in mind. But it is easy to show that x^4 is differentiable, as is x^3, and hence x^1/3 by the inverse function theorem, hence x^{4/3} is diffible, and it all follows some what simply from these big sounding theorems).
 
  • #5
...we haven't touched those theorems yet.

but, [tex]a=\left(8+h\right)^\frac{1}{3}[/tex] and [tex]b=\left(8\right)^\frac{1}{3}[/tex]
 
  • #6
x^n is differentiable at any "x", no matter the value of "n>0". The derivative is equal to n x^(n-1). Just plug n=4/3 and x=8 to see what you get.
 
  • #7
dextercioby said:
x^n is differentiable at any "x", no matter the value of "n>0".

I think he's been asked to show that, at least for n= 4/3.
 
  • #8
dextercioby said:
x^n is differentiable at any "x", no matter the value of "n>0". The derivative is equal to n x^(n-1). Just plug n=4/3 and x=8 to see what you get.

I see it equals what i wrote using l'hospitals rule, but through what derivative what did you go through to determine that?
 

What is L'Hospital's rule?

L'Hospital's rule is a mathematical technique used to evaluate limits of indeterminate forms in calculus. It states that for certain types of functions, the limit of the ratio of their derivatives is equal to the limit of the original function.

When is it appropriate to use L'Hospital's rule?

L'Hospital's rule is typically used when evaluating limits of the form 0/0 or ∞/∞. It can also be used when the limit is in an indeterminate form such as 1^∞, ∞^0, or 0^0.

How often can L'Hospital's rule be used?

L'Hospital's rule should only be used when all other methods of evaluating a limit have failed. It is not meant to be used as a shortcut and should be used sparingly.

Can L'Hospital's rule be used for all types of functions?

No, L'Hospital's rule can only be used for certain types of functions such as rational functions, exponential functions, and logarithmic functions. It cannot be used for trigonometric functions, inverse trigonometric functions, or piecewise-defined functions.

Are there any limitations to using L'Hospital's rule?

Yes, L'Hospital's rule may not always provide an accurate result. It can sometimes lead to incorrect solutions or infinite loops. It should always be used with caution and double-checked with other methods if possible.

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