Answer
We will call $x^{2}cos(20\pi x)$ is $g(x)$. We know that $-1\leq cos(x)\leq 1$, so do the same as $g(x)$, we have:
$x^{2}.(-1)\leq g(x) =x^{2}cos(20\pi x)\leq x^{2}.1$
According to Squeeze Theorem; we also have:
$\lim\limits_{x \to 0}[x^{2}.(-1)]=0$ and $\lim\limits_{x \to 0}(x^{2}.1)=0$
Therefore, we can conclude that:
$\lim\limits_{x \to 0}g(x)=0$
Note: The picture concludes three graphes: The violet graph is $g(x) =x^{2}cos(20\pi x)$, the red one is $f(x)=-x^{2}$ and another one is $h(x)=x^{2}$. In $g(x)$ graph, I take $\pi$ = 3.14.
Work Step by Step
We will call $x^{2}cos(20\pi x)$ is $g(x)$. We know that $-1\leq cos(x)\leq 1$, so do the same as $g(x)$, we have:
$x^{2}.(-1)\leq g(x) =x^{2}cos(20\pi x)\leq x^{2}.1$
According to Squeeze Theorem; we also have:
$\lim\limits_{x \to 0}[x^{2}.(-1)]=0$ and $\lim\limits_{x \to 0}(x^{2}.1)=0$
Therefore, we can conclude that:
$\lim\limits_{x \to 0}g(x)=0$
Note: The picture concludes three graphes: The violet graph is $g(x) =x^{2}cos(20\pi x)$, the red one is $f(x)=-x^{2}$ and another one is $h(x)=x^{2}$. In $g(x)$ graph, I take $\pi$ = 3.14.
