0
$$infty $$
1
-1
Answer is Right!
Answer is Wrong!
The correct answer is $\boxed{1}$.
The limit $\lim_{x\to 0}\frac{\sin^2(x)}{x}$ can be evaluated using the following steps:
- Substitute $x=0$ into the expression $\frac{\sin^2(x)}{x}$. This results in the indeterminate form $\frac{0}{0}$.
- Use L’Hôpital’s rule, which states that if $\lim_{x\to a}\frac{f(x)}{g(x)}$ is equal to the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x\to a}\frac{f'(x)}{g'(x)}$ exists and is equal to the limit $\lim_{x\to a}\frac{f(x)}{g(x)}$.
- Applying L’Hôpital’s rule to the limit $\lim_{x\to 0}\frac{\sin^2(x)}{x}$, we get $\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{1}$.
- Simplifying the expression $\lim_{x\to 0}\frac{2\sin(x)\cos(x)}{1}$, we get $\lim_{x\to 0}\sin(2x)$.
- Using the substitution $u=2x$, we get $\lim_{x\to 0}\sin(2x)=\lim_{u\to 0}\sin(u)$.
- The value of $\lim_{u\to 0}\sin(u)$ is equal to $1$.
- Therefore, the value of $\lim_{x\to 0}\frac{\sin^2(x)}{x}$ is also equal to $1$.
The other options are incorrect because they do not represent the value of the limit $\lim_{x\to 0}\frac{\sin^2(x)}{x}$.