Basic limit question ( f(x) = x^2 )

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In summary, in order to show that f(x) = x^2 approaches a^2 as x approaches a, we need to ensure that |x^2 - a^2| is less than a certain value (epsilon). This is achieved by bounding |x - a| and |x + a| separately, and then choosing the minimum of 1 and epsilon over (2|a| + 1) to ensure that both conditions are met. If either bound is unbounded, then the whole expression will not be bounded and the proof will not hold.
  • #1
ltjrpliskin
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f(x) = x^2 and we want to show that as x -> a, f(x) -> a^2.

So, we must ensure that |x^2 - a^2| < eps.

The book (Calculus by Spivak) starts of by bounding |x - a| < 1 which leads us to
|x + a| < 2|a| + 1.

This is the part where I get lost:
I can understand that we require |x - a| < eps/ (2|a| + 1),
but
why do we choose |x - a| < min(1, eps/ (2|a| + 1)) ?

And what happens if min(1, eps/ (2|a| + 1)) = 1 ?

Thank you very much for your help in advance! :)
 
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  • #2
hi ltjrpliskin! :smile:

(try using the X2 button just above the Reply box :wink:)

|x2 - a2| = |x - a||x + a|,

so we need |x + a| to be bounded

any bound would do, but it seems most sensible to use 1 :wink:
 
  • #3
tiny-tim said:
hi ltjrpliskin! :smile:

(try using the X2 button just above the Reply box :wink:)

|x2 - a2| = |x - a||x + a|,

so we need |x + a| to be bounded

any bound would do, but it seems most sensible to use 1 :wink:

Okay, but can you help me understand why we chose the minimum of 1 and eps/ (2|a| + 1)? :)
 
  • #4
we want |x - a||x + a| to be small …

|x - a| < 1 keeps |x + a| small, in fact < (2|a| + 1)

|x - a| < ε/ (2|a| + 1), keeps |x - a| small,

and both together keep the whole thing < ε :wink:

so we need both conditions … it won't work if |x - a| can be unbounded
 
  • #5
tiny-tim said:
we want |x - a||x + a| to be small …

|x - a| < 1 keeps |x + a| small, in fact < (2|a| + 1)

|x - a| < ε/ (2|a| + 1), keeps |x - a| small,

and both together keep the whole thing < ε :wink:

so we need both conditions … it won't work if |x - a| can be unbounded

Ahh! Okay, thank you very much :)
That was much simpler than expected haha :)
 

Related to Basic limit question ( f(x) = x^2 )

What is a basic limit question?

A basic limit question refers to a mathematical concept where the value of a function is approached as the input variable gets closer and closer to a specific value. It helps determine the behavior of a function at a certain point.

What is the function f(x) = x^2?

The function f(x) = x^2 is a polynomial function where the output (y) is equal to the input (x) squared. It is a simple and commonly used function in mathematics.

How do you find the limit of f(x) = x^2?

To find the limit of f(x) = x^2, we substitute the value that the input variable is approaching into the function. In this case, if the input variable is approaching a value of 'a', the limit would be written as lim x->a f(x) = a^2.

What is the value of the limit of f(x) = x^2 as x approaches infinity?

The value of the limit of f(x) = x^2 as x approaches infinity is infinity. This means that as the input variable increases without bound, the output value also increases without bound.

How is a basic limit question useful in real life?

Basic limit questions are useful in real life applications such as determining the maximum height a projectile can reach, calculating the speed of a moving object at a specific time, and analyzing the stability of a system. It is also used in various fields of science, engineering, and economics.

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