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For the function $f$, given by$$f(x) = \begin{cases} a & (x=0) \\ \frac {\sinh(x)}{\sinh(2x)} & (x\neq0)\end{cases}$$

Provided the limits exist, I need to determine, $$\lim_{x \to 0}f(x)$$

I know that $f(x)$ is undefined at $x=0$. Furthermore, I've determined that $\lim_{x \to 0} \frac {\sinh(x)}{\sinh(2x)} = \frac{1}{2}$.

Now, as far as my confusion is concerned. If the question only asked for $\lim_{x \to 0} \frac {\sinh(x)}{\sinh(2x)}$, I would simply show how I computed $\frac{1}{2}$.

However, with the additional case where $f(x) = a$, alongside the one-sided curly bracket notation, I'm not sure how to approach this problem. I cannot find any literature to enlighten me on how to use this notation in the context of limits, so I apologise if this is a duplicate. Perhaps this is because I don't know what name to give this one-sided curly bracket notation.

I do however have one idea on how to approach this:

Which is to show that $$\lim_{x \to 0} \frac {\sinh(x)}{\sinh(2x)} = \frac{1}{2} =a$$

Am I correct in my interpretation on how to handle this notation in the context of computing limits? Or am I missing something? Either way, how can this notation be explained in precise terms in this context? At the moment, it is fairly vague in my mind.

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    $\begingroup$ "one sided curly bracket notation": $f$ is a piecewise function $\endgroup$ – A. Goodier Nov 26 '17 at 15:41
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Limit of $f(x)$ when $x \to 0$ is about a neighborhood of $0$ which excludes zero itself.

You also mentioned

I know that f(x) is undefined at x=0

No, according to the definition $$f(0)=a$$

Limit is not about $f(0)$ at all. Even if $f(x=0)$ is undefined, it does not mean that this limit does not exists.

When saying

$$\lim _{x\to c}f(x)=L\iff (\forall \varepsilon >0,\,\exists \ \delta >0,\,\forall x\in D,\,0<|x-c|<\delta \ \Rightarrow \ |f(x)-L|<\varepsilon )$$

This definition clearly means $$x\ne c$$

The only case when it is important is when they state that $f$ is continuous at $0$. Then it means its limit and its value are equal leading to $a=\frac12$.

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  • $\begingroup$ Although it says f(0) = a, I can see that sinh(0)/sinh(0) = 0/0 which is undefined. Perhaps this is where my understanding of the notation is failing me? Are the two conditions mutually exclusive? So is this question only asking me to determine the limit for sinh(x)/sinh(2x) as x tends to 0? Note, that in the follow up question, I'm asked to compute values for a, for which f is continuous. So I guess this ties into your latter point? $\endgroup$ – HumptyDumpty Nov 26 '17 at 15:32
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    $\begingroup$ according to the notation, $sinh(x)/sinh(2x)$ is used only for when $x \ne 0$. Therefore, there is no mistake in the notation. Even everything is absolutely defined. Yes the conditions are mutually exclusive. If f(x) is continuous, then yes, a=1/2. But if there is no continuity condition imposed, a can be any arbitrary value. $\endgroup$ – Arash Nov 26 '17 at 15:36
  • $\begingroup$ Thank you for the clarification. Just to further clarify, can a limit at a be computed? Or given the fact that a limit for x tends to 0 is required, and since a is at x=0, maybe this isn't allowed, by the definition you've given? Since x would be tending to 0, where x is defined to be 0, so it would be 0 tends to 0? Or maybe the limit would be a itself in this case? I'm not sure. $\endgroup$ – HumptyDumpty Nov 26 '17 at 16:18
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    $\begingroup$ Limit at a? Do you mean limit at 0? Because a is a kind of y and not an x. This is probably not what you are looking for. Please consider that $x=0$ and $x\to 0$ are exclusive. It means that when $x\to 0$ then x is not zero $x\ne0$. The limit will be $1/2$ not a necessarily. $\endgroup$ – Arash Nov 26 '17 at 16:25

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