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

In summary: So I guess what I did in the first post doesn't make sense?In summary, to solve this problem, we can use the hint provided and let h=x-1. This allows us to rewrite the limit as \lim_{h \to 0} \frac{(1+h)^5-1}{h}. We can then compare this to the definition of a derivative, f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}, and see that it is equivalent to evaluating the derivative of f(x)=x^5 at x=1. Therefore, the solution to this problem is 5.
  • #1
swiftleaf
3
0
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?
 
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  • #2
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
gabbagabbahey said:
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
swiftleaf said:
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
gabbagabbahey said:
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.
 

1. What is the purpose of finding the limit by evaluating the derivative of a function?

The purpose of finding the limit by evaluating the derivative of a function is to determine the instantaneous rate of change of the function at a given point. This can be useful in various applications, such as finding the maximum or minimum value of a function, or determining the slope of a curve at a specific point.

2. How do you evaluate the derivative of a function to find its limit?

To evaluate the derivative of a function, you can use the limit definition of a derivative, which involves taking the limit of the difference quotient as the change in the input variable approaches zero. Alternatively, you can use differentiation rules, such as the power rule, product rule, or chain rule, to find the derivative and then take the limit of the resulting expression.

3. What is the difference between finding the limit of a function and finding the limit by evaluating the derivative?

The limit of a function is the value that the function approaches as the input variable approaches a certain value. On the other hand, finding the limit by evaluating the derivative involves determining the instantaneous rate of change of the function at a specific point. The former gives a general idea of how the function behaves, while the latter gives a more precise understanding of the function at a particular point.

4. Can finding the limit by evaluating the derivative be used for all types of functions?

Yes, finding the limit by evaluating the derivative can be used for all types of functions, including polynomial, exponential, trigonometric, and logarithmic functions. However, the process of finding the derivative may differ depending on the type of function.

5. Are there any limitations to finding the limit by evaluating the derivative?

One limitation is that this method may not work for functions that are not continuous at the point of interest. In such cases, other methods, such as using the limit definition of a derivative or graphical analysis, may need to be employed. Additionally, finding the derivative and taking the limit may be computationally challenging for some functions.

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