VCE Methods Question Thread!

[cosine]

Yes that is correct!

Alright, sounds good. Thanks 

[TrueTears]

You substitute whatever value 'x' is approaching to represent 'x' in 'x + 3'. In this case x is approaching 0.

Let's look at the graph y = x. If x approaches 5, then so will y. Since there is the coordinate (5, 5) on a y = x graph.

Hence, 0 + 3 = 3.

Why?

Consider the function , what is ?

[TrueTears]

Oh alright I get it now, thank you guys!

So the reason why we had to factor out the x was because if we didn't, we would have an undefined result as 0 as the denominator does not exist, right? And limit x------> 3 means to find the limit or f(x) value when x is approaching 3, right?


Thank you for the help!

Not quite.

Hint: think about the continuity at .

[Maths Forever]

Why?

Consider the function , what is ?

This only applies if 'x' is isolated, as in the case, x + 3. In your example 'x sin (1/x)' it is not possible to substitute 'x' straight away. This would require the 'sandwich theorem' which is not part of the VCE syllabus.

[TrueTears]

This only applies if 'x' is isolated, as in the case, x + 3.

This does not make sense.

In your example 'x sin (1/x)' it is not possible to substitute 'x' straight away.

Why can we 'substitute' into then? [The answer is not because its defined.]

Consider the following example:


What is ? By your approach, we can 'substitute' into , hence .

[Maths Forever]

This does not make sense.
Why can we 'substitute' into then? [The answer is not because its defined.]

Consider the following example:


What is ? By your approach, we can 'substitute' into , hence .

This is the limit law: Lim as 'x' approaches 'b' of 'x' = 'b'.

-1 ≤ sin(1/x) ≤ 1

Therefore -x ≤ x sin(1/x) ≤ x

Now by sandwich theorem, if Limit as x approaches 0 of '-x' is equal to the Limit as x approaches 0 of 'x', then Limit as x approaches 0 of 'x sin (1/x)' will exist.

Lim as x approaches 0 of '-x' = -(0) = 0

Lim as x approaches 0 of 'x' = 0 (by the limit law described above.

Hence the limit as x approaches 0 of 'x sin (1/x)' is 0. No need for the hybrid function.

By isolated, I meant 'x' is not part of a product function, like 'x sin (1/x)', not in the sense you described.

[TrueTears]

This is the limit law: Lim as 'x' approaches 'b' of 'x' = 'b'.

-1 ≤ sin(1/x) ≤ 1

Therefore -x ≤ x sin(1/x) ≤ x

Now by sandwich theorem, if Limit as x approaches 0 of '-x' is equal to the Limit as x approaches 0 of 'x', then Limit as x approaches 0 of 'x sin (1/x)' will exist.

Lim as x approaches 0 of '-x' = -(0) = 0

Lim as x approaches 0 of 'x' = 0 (by the limit law described above.

Hence the limit as x approaches 0 of 'x sin (1/x)' is 0. No need for the hybrid function.

My example above is completely different.

By isolated, I meant 'x' is not part of a product function, like 'x sin (1/x)', not in the sense you described.

See above example.

[Maths Forever]

My example above is completely different.

Sorry, what are you asking? Why do you think we can substitute 'x = 0' into 'x + 3'?

[TrueTears]

Sorry, what are you asking? Why do you think we can substitute 'x = 0' into 'x + 3'?

Yes, I am asking you to justify why you can 'substitute' into to find .

From what I can gather so far, as long as the function is 'defined' at that point and the function is not a product function ('isolated?'), then we can substitute in the value to find the limit. Hence, why I made the above (second) example. The function is clearly not a product function, is also well-defined at , hence using your approach, we can simply 'substitute' in into to find . This yields the (fallacious) answer of .

[Maths Forever]

You referred to 'x sin (1/x)' in your first message. So that is why I showed the limit. I know that g(x) is a separate example, and not a product function. The limit as x approaches 0 for g(x) is undefined. So you cannot substitute 'x=0' in this case and expect the limit to be 0, like you say. I was assisting with the limit as x approaches 0 for x(x+3) / x.

"From what I can gather so far, as long as the function is 'defined' at that point and the function is not a product function, then we can substitute in the value to find the limit."

We are in agreement for this. As x approaches 0 for x + 3, the function is defined.

[TrueTears]

The limit as x approaches 0 for g(x) is undefined. So you cannot substitute 'x=0' in this case and expect the limit to be 0.

There is a fallacy in this statement. You are assuming you already know the limit (which in this case is undefined), but how do you know the limit is undefined without computing the limit? That is why I am asking you to justify your approach to computing the limit.

"From what I can gather so far, as long as the function is 'defined' at that point and the function is not a product function, then we can substitute in the value to find the limit."

We are in agreement for this. As x approaches 0 for x = 3, the function is defined.

Ok, but the same conditions applies for my example:

1) is not a product function.
2) is defined at .

Hence, by your approach, , which is clearly false.

[Maths Forever]

How would you prove the limit as x approaches 0 for sin (1/x) is undefined (without your hybrid function)?

If at x = 0, g(x) = 0, there is no limit to evaluate.

Limit as x approaches 0 of '0' = 0. By the limit law, limit as x approaches 'b' of 'c' = 'c'. Do you know how to derive this?

[TrueTears]

How would you prove the limit as x approaches 0 for sin (1/x) is undefined (without your hybrid function)?

My hybrid function example has nothing to do with whether is undefined or defined as , it was to illustrate a fallacy in your approach.

If at x = 0, g(x) = 0, there is no limit to evaluate.

Yes there is.

Limit as x approaches 0 of '0' = 0. By the limit law, limit as x approaches 'b' of 'c' = 'c'. Do you know how to derive this?

Yes I do, but this has nothing to do with my example.

Perhaps an even simpler example to illustrate why your approach is incorrect is as follows. Define:



Again I ask, what is ? Using your approach:

1) g(x) is not a product function.
2) g(x) is defined at x=0.

Hence .

Also the above example illustrates why this statement is false.

If at x = 0, g(x) = 0, there is no limit to evaluate.

Clearly at , but there is an limit to evaluate, i.e., , whether it exists or not is not of the concern.

[lzxnl]

TT is getting at this point:

You have assumed that a limit of a function as x approaches b must be the function evaluated at that point if it exists. This is only actually true for a certain class of function called 'continuous functions'. It so happens that all polynomials, exponentials, logs, trig functions and sums/products of these are continuous over their maximal domains and so are quotients, assuming the denominator isn't 0. However, this is something that needs to be proved (not in VCE) and you can't just assume that the limit of x as x approaches 1 is 1. You need a rigorous working definition of a limit and then you need to work with it.

Fortunately this isn't covered in Methods so you don't have to worry about it. But at the same time, it's not healthy for your maths learning to just parrot limit theorems and calculations without understanding what's going on with them. Find a rigorous proof of the Sandwich theorem, or indeed l'Hopital's rule (which I'm sure you've seen); that'll be quite useful for your learning.

[TrueTears]

The point I'm trying to make is that you can only 'substitute' in x=0 into x+3 because the function g(x) = x+3 is continuous at x=0. Infact, let's just consider the statement . The issue of what happens when is completely irrelevant from the point of view of functional limits, actually doesn't even need to be in the domain of . Hence, why for my latest example, even though g(x) = 0 at x=0, this has no impact at all on the actual limit itself and simply substituting in x=0 into x+3 to find the limit would be completely wrong because the way I defined g(x) makes it such that it is not continuous anymore at x=0.

Thus, it is paramount you understand the (mathematical) definition and not just a conceptual understanding from reading "limit laws".