Why are these expressions indeterminate expressions? Why are these  $1^\infty,$ $0\cdot\infty$ and $\infty^0$ indeterminate forms. Why we can't solve these expressions? 
 A: Do you buy that $\frac{\infty}{\infty}$ and $\frac{0}{0}$ are indeterminate?  If so, it's easy to see why those are indeterminate:
\begin{align}
0\cdot\infty =& \frac{1}{\infty} \cdot \infty = \frac{\infty}{\infty} \\
0\cdot\infty =& 0 \cdot \frac{1}{0} = \frac{0}{0} \\
1^\infty =& e^{\ln(1)\cdot\infty} = e^{0\cdot\infty}\text{, the exponent is indeterminate} \\
\infty^0 =& e^{\ln(\infty)\cdot0} = e^{\infty\cdot0}\text{, again, the exponent is indeterminate}
\end{align}
Here is a way to informally see why they would be indeterminate.  You have to remember that these are really limits.  So we don't actually plug in $0$ and $\infty$.

*

*
*

*$\lim\limits_{x\rightarrow \infty}0\cdot x = 0$

*$\lim\limits_{x\rightarrow 0}x\cdot \infty = ?$ What is it approaching? At every point $x > 0$ this is infinite and at every point $x < 0$ it's $-\infty$.  So which is it? $0$ or $\pm\infty$?  It's indeterminate and it could be none of those.


*
*

*$\lim\limits_{x\rightarrow \infty} 1^x = 1$

*$\lim\limits_{x\rightarrow 1}x^\infty = ?$.  Again, when $x > 1$ this is infinite and when $x < 1$ (but very close to $1$, so still positive) this is $0$.  Again, so which is it? $0$, $1$, or $\infty$?  It's indeterminate and, again, could be none of those.


*
*

*$\lim\limits_{x\rightarrow \infty} x^0 = 1$

*$\lim\limits_{x\rightarrow 0}\infty^x = ?$.  Now when $x > 0$ this is infinite, when $x < 0$ this is $0$.  So, for the last time, which is it, $1$, $0$, or $\infty$--it's indeterminate.


A: You can't have cohesion if you accept a certain value for $0\cdot\infty$.
For example consider the expression $\lim\limits_{x\rightarrow \infty}\frac{k}{x}\cdot x= k$.
As you can see it depends on the value k. Similar arguments can be made for the other indeterminate forms you pose.
