How to solve this?
$\displaystyle \lim_{h \to 0}\left(0-h\right)^2sin\dfrac{1}{h}$

Hint: We are given to solve a limit. For this we should be aware about several formulae related to the limit of sine functions. We will first solve the bracket involving the variable $h$. Then we will convert this into a standard limit of some form and after that we will put the value of $h$ as equal to 0 after we are able to remove it from the denominator.

Complete step-by-step solution:
We have $\displaystyle \lim_{h \to 0}\left(0-h\right)^2sin\dfrac{1}{h}$
We will first solve $\left(0-h\right)^2=h^2$
Now, we know the fact that:
$\displaystyle \lim_{h \to 0}h\times sin\dfrac{1}{h}=0$
We will use this fact here and calculate the result accordingly.
We write $\displaystyle \lim_{h \to 0}h^2sin\dfrac{1}{h}$ as $\displaystyle \lim_{h \to 0}h\times h\times sin\dfrac{1}{h}$
Now, we separate the limit on the product using the following fact:
If $f$ and $g$ are two continuous functions of $x$ then the following holds true for all $x$:
$\displaystyle \lim_{h \to a}f\left(x\right)\times g\left(x\right)=\displaystyle \lim_{h \to a}f\left(x\right)\times \displaystyle \lim_{h \to a}g\left(x\right)$
So, we have:
$\displaystyle \lim_{h \to 0}h^2sin\dfrac{1}{h}=\displaystyle \lim_{h \to 0}h\times \left(\displaystyle \lim_{h \to 0} h\times sin\dfrac{1}{h}\right)$
$=0\times 0=0$
Using the fact that $\displaystyle \lim_{h \to 0}h=0$
$\displaystyle \lim_{h \to 0}\left(0-h\right)^2sin\dfrac{1}{h}=0$
Hence, the answer has been found out to be 0.

Note: You can directly use the formula that for any power of $n$ higher than 1 the result of $x^n sin\dfrac{1}{x}$ when $x\to 0$ is 0. Here n=2 which is greater than 1 hence we can directly write that the limit is 0. You should take care that if the limit is tending to 0 then the limit variable should not be present in the denominator. Always try to remove the variable in the denominator using formulae like used here and then put the limit.