Not clear to me why the limit as $x$ goes to $4$ of $\sqrt{x^2-16}$ is $0$, since the limits on both sides of $4$ are not the same. From the right it is zero, but from the left ($x= 3.99999$) is undefined.
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Using math expressions in posts is kind of neat; see intro to MathJax and $\LaTeX$. – hardmath May 06 '16 at 01:04
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You are correct, the limit s obviously not defined( for limit tending 4 or less than 4), but is defined ONLY if limit tends to greater than 4. – oops May 06 '16 at 01:04
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You may also draw a graph, to check the existence of limits and clarify your problem. – oops May 06 '16 at 01:07
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We can assert $\lim_{x \to 4} \sqrt{x^2-16} = 0$ unambiguously. Applying the definition of the limit to this, we are saying
$\forall \epsilon>0 \ \exists \delta>0$ such that $\forall x \in \text{dom}{(f)}$ satisfying $0<|x-4|<\delta$, $|\sqrt{x^2-16}|<\epsilon$.
In a small enough neighbourhood of $4$, there exist no $x$ satisfying the condition $x \in \text{dom}{(f)}$ to the left of $4$. We thus need only consider the right-hand side. The function is right-continuous at $x=4$ and so the condition is met.
If we didn't have this definition, there would be some not-so-nice consequences; one could restrict the domain so that otherwise continuous functions are not continuous at their endpoints.
MathematicsStudent1122
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In the context of a calculus (or pre-calculus) course, limits are often defined in terms of deleted neighborhoods of a point. Strictly speaking the definition of $\lim_{x\to a} f(x)$ does not depend on whether or not $f(x)$ is defined for $x=a$. Your interpolation of a restriction $\forall x \in dom(f)$ is found in few calculus/pre-calculus texts for this reason (assuming the real line and its metric topology as a given). – hardmath May 06 '16 at 04:12
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1@hardmath Well, I have the inequality $0<|x-a|$ in my answer, so is my definition also not in terms of deleted neighbourhoods?
I am saying that $\lim_{x \to a} f(x) = L$ means that for arbitrary $\epsilon$, $|f(x) - L|<\epsilon$ for all $x$ in the intersection of some deleted neighbourhood of $a$ and the domain of $f$. If you think there is anything inadequate with that, please elaborate.
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I don't have a calculus text that includes your restriction to the domain of the function. In my experience as a student and teacher, the issue of limits at an endpoint of an interval is treated with one-sided limits, so that the possibility of unequal left and right limits can be presented. Of course the OP has not cited their text. – hardmath May 06 '16 at 04:38
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1@hardmath The possibility of unequal left and right limits may still be presented. One just needs a function with a jump discontinuity; the limit will not exist by any definition, but the one-sided limits will. OP's example is slightly different; $f$ is undefined in $(4-\delta, 4)$. It is not a jump discontinuity. Regarding the issue of sources, I think most analysis (rather than calculus) books define it as I do. – MathematicsStudent1122 May 06 '16 at 04:59
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You are correct that we can only define the limit of $\sqrt{x^2-16}$ as $x$ approaches $4$ from above, precisely because the expression is undefined for $x \lt 4$.
The proper notation for this in most books is $\lim_{x\to4^+} \sqrt{x^2-16}=0$.
In a basic college calculus course, the student will likely see a definition of limit very similar to the following, quoted from the Univ. of Tenn. Math Archives Visual Calculus by Larry Husch:
Definition. The limit of $f(x)$ as $x$ approach $a$ is $L$ $$ \lim_{x\to a} f(x) = L $$ if and only if, given $\varepsilon \gt 0$, there exists $\delta \gt 0$ such that $0 \lt |x-a| \lt \delta$ implies that $|f(x)-L| \lt \varepsilon$.
This is somewhat casual in that no explicit requirement is made that $f(x)$ is defined anywhere. My understanding would be that $f(x)$ must at least be defined on the deleted neighborhood of $a$ given by $0 \lt |x-a| \lt \delta$. A Calculus I online tutorial does mention such a requirement:
Definition 1 Let $f(x)$ be a function defined on an interval that contains $x=a$, except possibly at $x=a$. Then we say that, $$ \lim_{x\to a} f(x) = L $$ if for every number $\varepsilon \gt 0$ there is some number $\delta \gt 0$ such that $$ |f(x)-L| \lt \varepsilon \;\; \text{ whenever } \;\; 0 \lt |x-a| \lt \delta $$
A pre-calculus text found through Google Books preview, Precalculus with Limits by Ron Larson, page 815, states the idea less formally:
Definition of Limit If $f(x)$ becomes arbitrarily close to a unique number $L$ as $x$ approaches $c$ from either side, then the limit of $f(x)$ as $x$ approaches $c$ is $L$. This is written as $$ \lim_{x\to c} f(x) = L. $$
In my view even this least formal statement requires an approach to the limit from both sides of the point $x=c$. Of course English being as ambiguous as it is, someone wanting to resist this requirement might interpret matters as only requiring one side or the other, not necessarily both. In mathematical practice, though, that weaker notion does not serve the purpose for which we want limits to be introduced, namely in order to define the derivative of a function at a point by taking the limit of the difference quotient.
hardmath
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To make it clear, let's look at the definition in your textbook. What is the title and author? – hardmath May 06 '16 at 14:15
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Okay, but there should be a definition of limit given in the notes. Is this a calculus or a pre-calculus class? – hardmath May 06 '16 at 19:07
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I realize that the answer online says zero, but it's just not clear to me why that is so if the definition is that it must be defined on both sides of the limit – user163862 May 06 '16 at 20:13
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If your teacher has prepared the class notes and you cannot find a definition of limit there, it is reasonable to ask. – hardmath May 06 '16 at 20:16
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I'll include a definition from a typical calculus or pre-calculus text by editing my answer. – hardmath May 06 '16 at 20:17
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