Limit question to be done without using derivatives

[vcsharp2003]

Homework Statement

If $\displaystyle{\lim_{x \rightarrow 1} {\frac {f(x) -2} {x^2 -1} }} = \pi$ then what is the value of the following limit $\displaystyle{\lim_{x \rightarrow 1} {f(x)}}$ ?

Relevant Equations

Quotient rule of Limits $\displaystyle{\lim_{x \rightarrow a} {\frac {f(x)} {g(x)} }} = \frac {\displaystyle{\lim_{x \rightarrow a} {f(x)}}} {\displaystyle{\lim_{x \rightarrow a} {g(x)}}}$ provided $\displaystyle{\lim_{x \rightarrow a} {g(x)}} \neq 0$

I am confused by this question. If I try applying the theorem under Relevant Equations then it seems to me that the theorem cannot be applied since the limit of the denominator is zero. This question needs to be done without using derivatives since it appears in the Limits chapter, which precedes the chapter on Derivatives.

Am I correct in saying that the quotient theorem of limits under Relevant Equations cannot be applied here?

Limit of numerator is $\displaystyle{\lim_{x \rightarrow 1} {(f(x)-2)} } = f(1) -2$

Limit of denominator is $\displaystyle{\lim_{x \rightarrow 1} {(x^2 -1)} }= 1^2 -1 = 0$

Since limit of denominator is 0, so we cannot say anything about the limit of $\frac {numerator}{denominator}$?

 

[BvU]

Since limit of denominator is 0, so we cannot say anything about the limit of $\frac {numerator}{denominator}$?

$\displaystyle{\lim_{x \to 1} {f(x) -2} }= ?$

It seems to me that you don't make use of the knowledge that

$$\lim_{x \rightarrow 1} {\frac {f(x) -2} {x^2 -1} } = \pi$$

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[vcsharp2003]

It seems to me that you don't make use of the knowledge that

$$\lim_{x \rightarrow 1} {\frac {f(x) -2} {x^2 -1} } = \pi$$

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What limit theorem would I apply here? The only theorem that comes to mind is the quotient theorem which seems to not apply here.
Sorry, I don't know how to answer your question. I can say that $\displaystyle{\lim_{x \to 1} {f(x) -2} }= 0$ but it doesn't make sense to me, since then the original limit would be $\frac {0} {0}$.

 

[BvU]

since then the original limit would be $0\over 0$.

But it is (very crudely expressed) !

 

[vcsharp2003]

But it is (very crudely expressed) !

Why would we conclude that limit of numerator is 0? Probably, if it were non-zero then the original limit would be indeterminate rather than $\pi$, and therefore it must equal 0.

In your view can we apply quotient rule of limits here?

 

[BvU]

Why would we conclude that limit of numerator is 0? Probably, if it were non-zero then the original limit would be indeterminate rather than $\pi$, and therefore it must equal 0.

If the given limit is $\pi$ and the denominator goes to zero, then the numerator has to go to zero too. That's all there is to this exercise.

In your view can we apply quotient rule of limits here?

I'm too rusty to be perfect in the terminology, but for this one it is a definite yes!

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[BvU]

As a 'challenge' : what is $f'(1)$ ?

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[vcsharp2003]

As a 'challenge' : what is $f'(1)$ ?

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That is a good question. I will need to think about it. Perhaps use the fundamental definition of a derivative using limits.

 

[vcsharp2003]

I'm too rusty to be perfect in the terminology, but for this one it is a definite yes!

But limit of denominator is 0, so it shouldn't apply.

 

[BvU]

But limit of denominator is 0, so it shouldn't apply.

Good thing I added a disclaimer

So I'm back to

If the given limit is π and the denominator goes to zero, then the numerator has to go to zero too. That's all there is to this exercise.

Good enough for me as a physicist. A neater way to express it would be using $\varepsilon$ that can be taken arbitrarily small and showing that $|f(1+\varepsilon)-2|$ can be made smaller than any $\delta >0$ based on the given limit.

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[vcsharp2003]

As a 'challenge' : what is $f'(1)$ ?

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I think $f'(1) =0$ since we can apply L'Hospital's rule here as shown below.

Is there an easier way?

 

[vcsharp2003]

As a 'challenge' : what is $f'(1)$ ?

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I have another solution as below that uses definition of limits.

 

[BvU]

I liked post #8 better !
But L'Hopital still works: However, you start with $\pi$, not with 0. Therefore $f'(1) = 2\pi$
as you corrected yourself, I see. Well done !

 

[vcsharp2003]

I liked post #8 better !
But L'Hopital still works: However, you start with $\pi$, not with 0. Therefore $f'(1) = 2\pi$
as you corrected yourself, I see. Well done !

Post#11 uses basic definition of derivative rather than L'Hospital's rule, which might be a better solution.

 

[vcsharp2003]

As a 'challenge' : what is $f'(1)$ ?

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That's a genius question by you.

 

[vcsharp2003]

As a 'challenge' : what is $f'(1)$ ?

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Even $f''(1)$ should equal $2\pi$. By again applying L'Hospital's rule to the solution in post#11, we get the above result.

 

[Mark44]

This question needs to be done without using derivatives since it appears in the Limits chapter, which precedes the chapter on Derivatives.

Based on this restriction, which is also stated in the thread title, what's the justification for using L'Hopital's Rule, which uses derivatives?

 

[Mark44]

Why would we conclude that limit of numerator is 0? Probably, if it were non-zero then the original limit would be indeterminate rather than π, and therefore it must equal 0.

You have this backwards. If the limit of the numerator was zero, then the original expression would be indeterminate. If non-zero, the expression would be undefined.

 

[vcsharp2003]

Based on this restriction, which is also stated in the thread title, what's the justification for using L'Hopital's Rule, which uses derivatives?

Yes, that shouldn't have been used.

 

[vcsharp2003]

You have this backwards. If the limit of the numerator was zero, then the original expression would be indeterminate. If non-zero, the expression would be undefined.

Then, this problem would not have a value for the required limit of f(x) as x approaches 1.

 

[Mark44]

You have this backwards. If the limit of the numerator was zero, then the original expression would be indeterminate. If non-zero, the expression would be undefined.

Then, this problem would not have a value for the required limit of f(x) as x approaches 1.

You still have things backwards. Clearly, the limit of the denominator is zero. If the limit of the numerator also is zero, then the expression is indeterminate. That's the only way that the limit of the overall expression could turn out to be $\pi$. If the limit of the numerator happened to be anything other than zero, then the limit of the original expression would not exist.

Since $\lim_{x \to 1} f(x) = 2$, then f(1) = 2.

 

[vcsharp2003]

You still have things backwards. Clearly, the limit of the denominator is zero. If the limit of the numerator also is zero, then the expression is indeterminate. That's the only way that the limit of the overall expression could turn out to be $\pi$. If the limit of the numerator happened to be anything other than zero, then the limit of the original expression would not exist.

Since $\lim_{x \to 1} f(x) = 2$, then f(1) = 2.

Yes, isn't that what I said? In my post#5 I said as in quotes below. I reasoned that the numerator must be zero for the limit to have a value of $\pi$.

"Why would we conclude that limit of numerator is 0? Probably, if it were non-zero then the original limit would be indeterminate rather than π, and therefore it must equal 0."

 

[vcsharp2003]

You have this backwards. If the limit of the numerator was zero, then the original expression would be indeterminate. If non-zero, the expression would be undefined.

Oh, I see what you're saying. I thought undefined is same as indeterminate. They are very different, which I learned from your post. Thank you.

 

[vela]

But limit of denominator is 0, so it shouldn't apply.

You could do something like this.
$$\lim_{x\to 1}\frac{f(x)-2}{x+1} = \lim_{x\to 1} \frac{f(x)-2}{x^2-1} \lim_{x \to 1} (x-1) = 0.$$ Then use the quotient rule for limits on the first limit.

 

[Mark44]

Oh, I see what you're saying. I thought undefined is same as indeterminate. They are very different, which I learned from your post.

Yes, that was my point.

 

[vcsharp2003]

You could do something like this.
$$\lim_{x\to 1}\frac{f(x)-2}{x+1} = \lim_{x\to 1} \frac{f(x)-2}{x^2-1} \lim_{x \to 1} (x-1) = 0.$$ Then use the quotient rule for limits on the first limit.

Wow, that's a very intelligent solution. I never thought about this approach.