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I was reading Calculus by Michael Spivak, when I came across this problem given in the textbook..
Compute the limit $$\lim_{x\to 2}{\frac{x^3-8}{x-2}}$$ Now, this is a pretty standard limit and I know there are a couple of ways to solve it..by factoring or by using L Hospital’s rule...but here..I want to solve the limit by factoring..

My question is, when we solve the limit by factoring..we make an assumption that $x$ is not equal to 2. Now, yea I know that $x$ is only approaching the value 2..and hence the assumption is justified...but then this is not what the definition says right?

To me, the definition of a limit allows us to ignore the value 2 ,while verifying that a certain number is the limit of the function or not..by finding a $\delta$ given an $\epsilon$.

What I mean to say is..the word “approach” is quite subjective and to me it’s not very convincing that the limit will be the same even if we ignore the value 2...so I am looking for a rigorous line of reasoning entirely based on the definition of a limit which would then allow us to completely ignore the value of the function at $x=2$ even while COMPUTING the limit...not just while VERIFYING whether a certain number is the limit or not

PS: My question is a little “dry”, in the sense it asks for something which should be obvious..but I’m still confused..because the definition of the limit is something very fundamental..and any such assumption made while computing limits must follow from the definition..

EDIT: I have tried to explain my question in a better way here

Thanks for any answers!!

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    $\begingroup$ By definition of the limit the value of the function considered is never evaluated at the limit point. Also the function provided is clearly not defined at $x=2$ so why would it matter? $\endgroup$ – Peter Foreman May 7 at 9:46
  • $\begingroup$ Yes..as I said..we do ignore the limit point while verifying whether a certain number is the limit or not..the definition specifically asks us to do that by saying $0<|x-a|<\delta$...but how do we justify the same while evaluating the limit? $\endgroup$ – thornsword May 7 at 9:49
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The definition of a limit does not include the words like "approaches" or "tends" in a semantic manner but rather as a part of notation. The definition just asserts the truth of a complicated logical statement.

Further the definition of limit can not be used directly to evaluate limit of a function but it can be used to check if a given number is limit of function or not (and even this part requires some experience). The power of definition of limit comes from the fact that it can be used to prove general theorems which can be used to evaluate limits very effectively.

Coming to your problem at hand you have option to use the definition directly but then you need to guess a value for the limit and then check whether it is indeed the limit or not. The desired limit is $12$ and you should try to check this via definition of limit.

Another option is to use limit laws as explained in this answer.

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Limits are essentially used

  • when a function is undefined at some point but not in its neighborhood, and we want to continue it,

  • when we want to check continuity, by comparing the limit and the function value.

In the first case, we have no function value at the given point, so your question is pointless.

In the second case, we must not use the known function value, and your question is answered.

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Notice the the original function is not defined at $2$ in the first place, and the limit is only concerned with values infinitely close to $2$ (but never equal to $2$). The left and right limits will approach whatever the factored version of the function at $2$ would equal.

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  • $\begingroup$ But how does this follow from the definition - "The left and right limits will approach whatever the factored version of the function at 2 would equal." ? $\endgroup$ – thornsword May 7 at 9:51
  • $\begingroup$ @binarybitarray The factored and unfactored versions of the function are the same, except for the fact that the latter is undefined at $x=2$. We are not interested in what happens at exactly $2$, the fact that the factored function at $2$ equals our limit is a consequence of its continuity. $\endgroup$ – Tavish May 7 at 10:04
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When you are using the formal definition, the assumption is for every \epsilon exists $\delta$ so that if $0<|(x-x_0)|<\delta$, then $|f(x)-f(x_0)|<\epsilon$. meaning you do not equal $x_0$ exactly but in a close proximity not containing the point itself. It is important mainly becuase of issues with "plugging in" the point $x_0$ itself as in examples that you have given (it is just invalid because the function is not even defined there).

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  • $\begingroup$ I agree! But how does the same line of reasoning work while evaluating limits? There is a distinction between verifying and actually computing a limit right?...So how can we ignore the limit point while evaluating the limit? $\endgroup$ – thornsword May 7 at 9:53
  • $\begingroup$ All methods of evaluation are based off the definition. When you made "leeways" so that you could compute it is always there because you are using the definition in the background, so if you do things that are not allowed by the formal rules you may achieve results which contradict the argument (like plugging $x=2$ above where it is not defined) $\endgroup$ – Orenio May 7 at 9:55
  • $\begingroup$ But...to me personally...this is not as rigorous as other theorems in math....so is there any other way of thinking about this? $\endgroup$ – thornsword May 7 at 10:02
  • $\begingroup$ You are trying to figure out whether there exists a $g(x)$ that satisfies $g(x)=f(x)$ for all $x$ other than 2, and at $x=2$ equals to the limit of $f(x)$ at 2.That function $g(x)$ will than be defined as the continuous expansion of f so to speak; $g(x)$ is continuous everywhere. So you may not plug in the specific $x$ initially, but you can with $g(x)$, and since both $f$ and $g$ are "the same", it's actually ok. $\endgroup$ – Orenio May 7 at 10:16

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