Find the following limits or state that they do not exist. \lim_{x\to1^+}

Use the definition of continuity and the properties of limits to show that the function
${\displaystyle f\left(x\right)=x\sqrt{16-x^2}}$ is continuous on the interval [-4,4]

Evaluate the following limits. $\underset{x\to 0^{+}}{lim}{x}^{x^2}$

How can I calculate the following difficult limit: ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(x+\sin x)\sin \frac{1}{x}}$ ?

Limit of a function as it tends to zero
${\displaystyle \underset{x\to 0}{lim}\frac{3{\left(\tan 2x\right)}^2}{2x^2}}$
What method is best for solving this? The division by zero is causing me issues when trying to apply Sandwich theorem and/or Algebra of limits.

Find the limit, is the function continuous at the point being approached
${\displaystyle \underset{x\to \frac{\pi }{6}}{lim}\sqrt{{\csc }^{2}x+5\sqrt{3}\tan x}}$

Limit ${\displaystyle \underset{lim}{lim}its\{x\to 0\}\left\{\frac{\tan \left\{x\right\}-\sin \left\{x\right\}}{x^3}\right\}}$
So what I tried is:
${\displaystyle \underset{x\to 0}{lim}\frac{\frac{\sin \left\{x\right\}}{\cos \left\{x\right\}}-\sin \left\{x\right\}}{x^3}=\underset{x\to 0}{lim}\frac{\left\{\sin \left\{x\right\}\right\}(\frac{1}{\cos x}-1)}{x\cdot x^2}}$
From here, using the rule ${\displaystyle \underset{x\to 0}{lim}\frac{\sin \left\{x\right\}}{x}=1}$ it remains to evaluate
${\displaystyle \underset{x\to 0}{lim}\frac{\frac{1}{\cos \left\{x\right\}}-1}{x^2}}$