Use L'Hôpital's, and a table of values and a graph to find the following limits.
\[
\lim _{t \rightarrow 0} \frac{\sin \left(t^{2}\right)}{t^{2}}
\]
\[
\lim _{x \rightarrow 0^{+}} x \ln x
\]

Step 1 ：Given the limit problems: \(\lim _{t \rightarrow 0} \frac{\sin \left(t^{2}\right)}{t^{2}}\) and \(\lim _{x \rightarrow 0^{+}} x \ln x\)

Step 2 ：For the first limit, it is of the form 0/0 when t approaches 0, so we can apply L'Hôpital's rule. L'Hôpital's rule states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives, provided the limit exists.

Step 3 ：Find the derivative of the numerator and the denominator of the first function. The derivative of the numerator \(\sin \left(t^{2}\right)\) is \(2t\cos \left(t^{2}\right)\) and the derivative of the denominator \(t^{2}\) is \(2t\).

Step 4 ：Find the limit of the quotient of these derivatives as t approaches 0. The limit is \(\frac{2t\cos \left(t^{2}\right)}{2t}\) which simplifies to 1.

Step 5 ：For the second limit, it is also of the form 0*infinity when x approaches 0 from the right. This is an indeterminate form, but it's not in the form that we can directly apply L'Hôpital's rule.

Step 6 ：We can rewrite the second function in the form 0/0 or infinity/infinity by taking the reciprocal of one of the factors. The function \(x\ln x\) can be rewritten as \(\frac{\ln x}{1/x}\).

Step 7 ：Find the derivative of the numerator and the denominator of the rewritten second function. The derivative of the numerator \(\ln x\) is \(1/x\) and the derivative of the denominator \(1/x\) is \(-1/x^{2}\).

Step 8 ：Find the limit of the quotient of these derivatives as x approaches 0 from the right. The limit is \(\frac{1/x}{-1/x^{2}}\) which simplifies to 0.

Step 9 ：Final Answer: The limit of the first function as t approaches 0 is \(\boxed{1}\) and the limit of the second function as x approaches 0 from the right is \(\boxed{0}\).