evaluate the limit: $lim_{x\rightarrow0}\frac{x^{2}}{sin(x^{2})}$

Explanation:

Step1: Recall the limit - formula

We know the well - known limit $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$. Let $u = \frac{1}{x}$. As $x\rightarrow\infty$, then $u=\frac{1}{x}\rightarrow0$.

Step2: Rewrite the given limit

The given limit is $\lim_{x\rightarrow\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}$. Substituting $u = \frac{1}{x}$, when $x\rightarrow\infty$, $u\rightarrow0$. So, $\lim_{x\rightarrow\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}=\lim_{u\rightarrow0}\frac{\sin u}{u}$.

Step3: Evaluate the limit

Since $\lim_{u\rightarrow0}\frac{\sin u}{u}=1$, then $\lim_{x\rightarrow\infty}\frac{\sin(\frac{1}{x})}{\frac{1}{x}} = 1$.

Answer:

1