How to show that a limit cannot be another number? Let:
$$
G(x) = \left\{ \begin{array} {cc} x \sin \frac{1}{x} , & x\neq 0 \\
0, & x=0 \end{array} \right.
$$
I can understand that the function is continuous at $x=0$ because:
For $\epsilon>0$ and $\delta>0$, this implies, for all $x$

$$
 |x-0|<\delta \implies |f(x)-0| < \epsilon \\
 |x|<\delta \implies |f(x)| < \epsilon \\
 |x|<\delta \implies |x\sin \frac{1}{x}| < \epsilon \\
|x|<\delta \implies |x||\sin \frac{1}{x}| < \epsilon \\
\therefore \delta = \frac{\epsilon}{\sin \frac{1}{x}}
$$
So, for any $\epsilon>0$, we can find a $\delta=\frac{\epsilon}{\sin \frac{1}{x}}$ so that $ |x-0|<\delta \implies |f(x)-0| < \epsilon$ is true.
Hence, $\lim \limits_{x\to 0}G(x)=G(0)=0$

My text also mentions that $G$ will only be continuos at $0$ if $G(0)=0$.
So now I am wondering why can't it be any number, $l$?
So far here is what I have come up with using the same manipulations as above:

$$|x-0|<\delta \implies |f(x)-l| < \epsilon \\
|x|< \frac{\epsilon + l}{\sin \frac{1}{x}}
$$

I had expected this to lead to a contradiction when $l \neq 0$ but so far I can't see it. How do I show that $\lim \limits_{x \to 0}G(x)$ must be $0$ and where did I go wrong in my workings?
Thank you in advance for any help provided.
 A: Hint Assume that for some $l$ the function 
$$H(x) = \left\{ \begin{array} {cc} x \sin \frac{1}{x} , & x\neq 0 \\
l, & x=0 \end{array} \right.$$
Then, the function 
$$H(x)-G(x) = \left\{ \begin{array} {cc}0 , & x\neq 0 \\
l, & x=0 \end{array} \right.$$
is continuous at $0$ as the difference of two continuous functions.
A: Your text is correct. The function is continuous only if we define $G(0)=0$. This is because to get continuity of $G$ at $0$ we need that $G(0)=\lim_{x\to0}G(x)$ and limits of functions at some point in $\mathbb{R}$ (provided the limit exists) are unique when we determine convergence by means of the $\epsilon-\delta$  method.
To see this suppose that $f$ is a function defined on a neighbourhood of $a$ such that $\lim_{x\to a}f(x)$ exists where $a\in\mathbb{R}$. I claim the limit is unique.
Suppose not. Then there are two distinct numbers $L,M\in\mathbb{R}$ such that $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}f(x)=M$. This means that if we take $0<\epsilon=\frac{|L-M|}{100}$ there exists $0<\delta_{1}$ and $0<\delta_{2}$ such that if $0<|x-a|<\delta_{1}$ then $|f(x)-L|<\epsilon$ and if $0<|x-a|<\delta_{2}$ then $|f(x)-M|<\epsilon$. If we take $\delta=\min\{\delta_{1},\delta_{2}\}$ then if $0<|x-a|<\delta$ then 
$|L-M|=|(L-f(x))+(f(x)-M)|\le|L-f(x)|+|f(x)-M|<\epsilon+\epsilon=2\epsilon=\frac{|L-M|}{50}$
by the definition we gave fo $\epsilon$. But this implies $49|L-M|<0$. But this is impossible. Contradiction.
Hence limits must be unique. So the reason that we can't define $G(0)$ as anything else is because if the limit exists then it is unique. Since $|x\sin\big(\frac{1}{x}\big)|\le|x|$ then $lim_{x\to0}G(x)$ exists and is $0$.
A: First of all, there is a caveat about your proof, the value you choose for $\delta$ can get very big or not even be a number, for example lets fix $x$ such that $\sin{\frac{1}{x}}=0$ (which actually happens an infinite number of times as x approaches to 0), then your 
$$
\delta=\frac{\varepsilon}{\sin{\frac{1}{x}}},
$$
is not well defined.
However, your proof is not all wrong you are just picking the wrong delta, in this case, as your function $G$ is multiplied by the well behaved identity function you can have $\delta=\varepsilon$, so the proof that $G$ is continuous at $x=0$ will be as follows:
Let $\varepsilon>0$ given and let $\delta=\varepsilon$, then for any $x$ such that:
$$
|x-0|<\delta
$$
we have that:
$$
|G(x)-G(0)|=|x\sin{\frac{1}{x}}-0|\\
           =|x\sin{\frac{1}{x}}|\\
           =|x||\sin{\frac{1}{x}}|\\
          \le|x|\\
           \lt\delta=\varepsilon
$$
where your missing step follows from the fact that $|sin(y)|\le1$ for all $y\in\mathbb{R}$, and the triangle inequality. So we conclude that $G$ is continuous at $x=0$. 
Now if you want to see why any other choice of $l$ will fail, your assumption of using a proof by contradiction is correct, nevertheless you are doing your algebra wrong.
Lets suppose there is an $l\not=0$ such that $G(0)=l$ is continuous at $x=0$. This means that for every $\varepsilon>0$ we can find a $\delta_\varepsilon>0$ (the subscript is just o make emphasis on the fact that the choice of this $\delta$ depends on the $\varepsilon$ given) such that:
$$
|x-0|<\delta_\varepsilon \Rightarrow |G(x)-G(0)|<\varepsilon.
$$
So the proof goes like this:
Let $\varepsilon>0$ given, and $\delta_\varepsilon$ such that:
$$
|x-0|<\delta_\varepsilon \Rightarrow |G(x)-G(0)|<\varepsilon.
$$
To make our life easier, lets take another $\delta$ to be: $\delta=\min\{\delta_\varepsilon,\varepsilon\}$, (Note that $|x-0|<\delta<\delta_\varepsilon$ so the conclusion of our assumption still holds.) hence:
$$
|x-0|<\delta,
$$
then:
$$
\left||G(x)|-|G(0)|\right|\le|G(x)-G(0)|<\varepsilon,
$$
so:
$$
\left||x\sin{\frac{1}{x}}|-|l|\right|<\varepsilon,
$$
it follows that:
$$
-\varepsilon<|x\sin{\frac{1}{x}}|-|l|<\varepsilon,
$$
in particular, using the left side:
$$
|l|-\varepsilon<|x\sin{\frac{1}{x}}|
$$
but remember that $|x\sin{\frac{1}{x}}|\le|x|$ so:
$$
|l|-\varepsilon<|x\sin{\frac{1}{x}}|\le|x|=\delta\le\varepsilon.
$$
so we conclude that:
$$
|l|<2\varepsilon.
$$
Which is clearly a contradiction because $l$ is fixed, and this should hold for any value of $\varepsilon$.
So $G$ is not continuous at $x=0$ if $G(0)\not=0$.
A: The direct question was not answered:

I had expected this to lead to a contradiction when $l \neq 0$ but so far I can't see it. How do I show that $\lim \limits_{x \to 0}G(x)$ must be $0$ and where did I go wrong in my workings?

You cannot get a contradiction because your workings are (with some minor re-arrangement) essentially correct.
You did not get a proof that the function with $G(0)=L$ is continuous, because for $L \neq 0$ those workings are proving something slightly different than continuity. 
The endpoint of your reasoning with inequalities would be a correct proof of the true statement that when $G(0)$ is assigned the value $L$, then: for all $\epsilon > 0$, there exists a $\delta > 0$ such that:

$|G(x)| < |L| + \epsilon \quad$ when $|x| < \delta$.

For continuity with $G(0)=L$, that phrase should be replaced by

$L - \epsilon < G(x) < L + \epsilon \quad$ when $|x| < \delta$.

If $L=0$ the two phrases have the same meaning, so what your argument would demonstrate is that the function with $G(0)=0$ is continuous.  For other values of $L$, it is a proof of the upper bound, but the lower bound needed to get a proof of continuity is false.  There is no contradiction, because there is no proof of the false statement.
The re-arrangements of the proof needed for full correctness are:


*

*replace sin by |sin| in the inequalities

*bound |sin| by $1$, as suggested by several people

*which leads to $\delta = \epsilon$ as the value to use in the proof

*present the inequalities starting in the opposite order, showing that if $\delta = \epsilon$, then the conditions for continuity (with $G(0)=0$) are satisfied.

