Will the capacity of a channel becomes unbounded if i increase its signal-to-noise ratio $S/N$ without limit? According to the Shannon-Hartley theorem the capacity $C$ of a channel which has a signal-to-noise ratio of $S/N$ and a bandwidth $B$ can be calculated to be $C = B \log_2 \left( 1 + \frac{S}{N}\right)$. 
Heren if $B→\infty$, the capacity does not become infinite since,
with an increase in bandwidth,the noise power also increases. If the noise
power spectral density is $N_0/2$,then the total noise  powr is $N=N_0B$,so the Shannon-Hartley law becomes
\begin{align}
C &amp;= B \; \log_2 \left( 1 + \frac{S}{N_0B}\right) \\
&amp;=\frac{S}{N_0}\left(\frac{N_oB}{S}\right)\log_2\left( 1 + \frac{S}{N_0B}\right) \\
&amp;=\frac{S}{N_0}\log_2\left( 1 + \frac{S}{N_0B}\right)^{(\frac{N_0B}{S})}.
\end{align}
Now
$$\lim_{x\to0}(1+x)^{1/x}=e$$
So now it becomes
$$C_∞=\lim_{B\to\infty} C=\frac{S}{N_0}\log_2e=1.44\frac{S}{N_0}$$
so here channel capacity does not become infinite; that means, it's bounded.

But what happens if we increase the signal to noise ratio without bound? Will that give unbounded capacity? Is it possible? 

After searching I got this but don't know how is this is possible; can some one justify this?
By the sampling theorem, uniform samples taken at more than twice the bandwidth of a band-limited signal can be used to reconstruct the signal perfectly by sinc interpolation. For each different set of sample values there is a different band-limited function. In absence of noise, by increasing the bit depth of the samples one can store an unlimited amount of information per sample into the band-limited signal, and get it back simply by sampling and digitizing it again.