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Find the given limit by evaluating the derivative of a suitable function

  • Thread starter swiftleaf
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This is calculus I by the way.


Homework Statement



Find the given limit by evaluating the derivative of a suitable function at an appropriate point.

lim x->1 (x^5-1)/(x-1)

Homework Equations


None really but they have hints:
Hints were: look at the definition of the derivative, and then also let h=x-1.


The Attempt at a Solution



I can solve this question two ways (L'hospital and regular factoring)
Factor (x^5-1), to get (x-1)(x^4+x^3+x^2+x+1).
The (x-1)'s cancel out and you are left with:
lim x->1 (x^4+x^3+x^2+x+1) = 5

Or L'hospital 5x^4 = 5

I really don't understand what method the question is asking me to do. I can try the definition of a derivative method but I get really stuck..

[f(x+h)-f(x)]/h

(assuming my function is (x^5-1)/(x-1))

[(x+h)^5-1]/(x+h)-(x^5-1)/(x-1)) / h

Then that just looks like a mess when I expand it.. I doubt this is the way they want us to do it.. anyone want to provide any insight on how to do this?
 

Answers and Replies

  • #2
gabbagabbahey
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Well, if you let [itex]h=x-1[/itex] as the hint suggests, then [itex]x=1+h[/itex], and

[tex]\lim_{x\to 1} \frac{x^5-1}{x-1} = \lim_{h \to 0} \frac{(1+h)^5-1}{h}[/tex]

Compare that to the definition of the derivative of a function [itex]f[/itex] at a point [itex]a[/itex]:

[tex]f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}[/tex]

Can you think of a suitable [itex]f[/itex] and [itex]a[/itex] to turn your limit into the definition of a derivative (Hint: [itex]1=1^5[/itex] :wink:)?
 
  • #3
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Well, if you let [itex]h=x-1[/itex] as the hint suggests, then [itex]x=1+h[/itex], and

[tex]\lim_{x\to 1} \frac{x^5-1}{x-1} = \lim_{h \to 0} \frac{(1+h)^5-1}{h}[/tex]

Compare that to the definition of the derivative of a function [itex]f[/itex] at a point [itex]a[/itex]:

[tex]f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}[/tex]

Can you think of a suitable [itex]f[/itex] and [itex]a[/itex] to turn your limit into the definition of a derivative (Hint: [itex]1=1^5[/itex] :wink:)?
Sort of. I can think of f(x)=x^5 which would satisfy the equation but then the expansion of (x+h)^5 would be ridiculously long
h^5+5 h^4 x+10 h^3 x^2+10 h^2 x^3+5 h x^4+x^5-x^5/h

Then I'd get h^5+5 h^4 x+10 h^3 x^2+10 h^2 x^3+5 h x^4/h

That's a long and tedious way to get the answer but it works I guess! Thanks for the help. I actually tried this at the beginning but I was just too lazy to factor out the (x+h)^5 xD. Is there any shorter way to do this question? Other than the methods I tried before?
 
  • #4
gabbagabbahey
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Sort of. I can think of f(x)=x^5 which would satisfy the equation but then the expansion of (x+h)^5 would be ridiculously long
You're not supposed to expand it, the problem says to evaluate the derivative, so what is the derivative of f(x)=x^5 at x=1?

I can understand your confusion, since it is certainly a backwards way of doing things, the fact that [itex]\frac{d}{dx}x^5=5x^4[/itex] ultimately comes from evaluating the limit, so using the derivative to evaluate the limit (rather than evaluating the limit to get the derivative) is an odd way of doing things. However, that's what the problem says to do, so...
 
  • #5
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You're not supposed to expand it, the problem says to evaluate the derivative, so what is the derivative of f(x)=x^5 at x=1?

I can understand your confusion, since it is certainly a backwards way of doing things, the fact that [itex]\frac{d}{dx}x^5=5x^4[/itex] ultimately comes from evaluating the limit, so using the derivative to evaluate the limit (rather than evaluating the limit to get the derivative) is an odd way of doing things. However, that's what the problem says to do, so...
Oh so basically the hints were to tell me to compare the definition of a derivative and to use x=h+1 in order to make the limit look like the definition of a derivative just so I can discover a function that can satisfy the limit?

Thanks for your clarification! I was just confused most of the time by what the question really meant xP I kept trying to use the definition of a derivative to solve instead of using it to just compare, but that just made things very complicated.
 

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