- #1
fishspawned
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Homework Statement
the limit x->a of [(x+2)^5/3 - (a+2)^5/3] / (x-a)
Homework Equations
limit laws
Limit as x approaches a: (x+2)^5/3 - (a+2)^5/3 / (x-a) | Limit Laws
In summary, the limit x->a of [(x+2)^5/3 - (a+2)^5/3] / (x-a) is homework equation. The Attempt at a Solution mentions that differentiation is the shortest way. The restriction is due to the question is apparently brought up before any derivatives are introduced. So apparently this is do-able without using straight up differentiation. The 1/3 is killing me. and i have no idea how to approach it.
the limit x->a of [(x+2)^5/3 - (a+2)^5/3] / (x-a)
limit laws
- #2
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Does the formula in the third attempt remind you on something?
- #3
fishspawned
- 66
- 16
it looks like I'm trying to get a derivative. I am trying to do this without working using any derivatives and for some reason the power of 5/3 is completely messing with my head and i do not know how to proceed. i am still getting a 0/0 situation.
- #4
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Well, differentiation seems to be the shortest way. Otherwise you probably will have to follow the paths the differentiating formulas are proven. The power ##\frac{1}{3}## is the difficulty here, for you cannot expand it easily. Why do you want to restrict yourself, once you already have the formula for the first derivative?
- #5
fishspawned
- 66
- 16
The restriction is due to the question is apparently brought up before any derivatives are introduced. So apparently this is do-able without using straight up differentiation. Yes, the 1/3 is killing me. and i have no idea how to approach it.
- #6
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I've just looked up how ##\frac{d}{dx} x^{\alpha}=\alpha x^{\alpha-1}## is proven for non-integer values ##\alpha##.
It's done by the chain rule and ##x^\alpha = \exp(\alpha \ln x)##. Perhaps this might help and you may use properties of the exponential function. Another substitution ##y=x+2## should decrease writing work.
It's done by the chain rule and ##x^\alpha = \exp(\alpha \ln x)##. Perhaps this might help and you may use properties of the exponential function. Another substitution ##y=x+2## should decrease writing work.
- #7
Delta2
Gold Member
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Set ##X=(x+2)^{1/3}## ##A=(a+2)^{1/3}## then the it becomes ##\lim_{X\rightarrow A}{\frac{X^5-A^5}{X^3-A^3}}## . Hope I am right here and this helps.
- #8
fishspawned
- 66
- 16
DELTA:
this seems to take care of it - provided you know how to expand a sum of powers to a high level - had to look it up to be honest:
so revised sheet:
this seems to take care of it - provided you know how to expand a sum of powers to a high level - had to look it up to be honest:
so revised sheet:
- #9
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Correct result.fishspawned said:DELTA:
this seems to take care of it - provided you know how to expand a sum of powers to a high level - had to look it up to be honest:
so revised sheet:
FAQ: Limit as x approaches a: (x+2)^5/3 - (a+2)^5/3 / (x-a) | Limit Laws
What is a limit?
A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a particular value. It represents the value that a function approaches, but may never actually reach, as the input gets closer and closer to a specific value.
How do you find a limit?
To find the limit of a function as its input approaches a particular value, you can use algebraic techniques, such as factoring or simplifying, or you can use graphical techniques, such as using a graphing calculator or graphing the function by hand. You can also use numerical methods, such as plugging in values close to the limit or using a calculator to calculate the function at different values.
What does it mean for a limit to exist?
A limit exists when the value that the function approaches as its input gets closer and closer to a particular value is equal to a single, finite value. This means that the function is well-defined and has a consistent value at that point, even if it is not defined at that exact point.
What is the difference between a left-sided limit and a right-sided limit?
A left-sided limit is the value that the function approaches as its input gets closer and closer to a particular value from the left-hand side. Similarly, a right-sided limit is the value that the function approaches from the right-hand side. In some cases, these two limits may be different, indicating that the function has a discontinuity at that point.
Why is it important to find limits?
Finding limits is important in calculus because it allows us to analyze the behavior of a function and make predictions about its values. Limits are also used to define other important concepts, such as derivatives and integrals, which are essential in various fields of science, engineering, and economics.