Using the Squeeze theorem, show that

Limit x-0 x^2cos(1/x)=0

Question

Using the Squeeze theorem, show that

Limit x->0         x^2cos(1/x)=0

anonymous · Answer

Thankfully, the range of $\cos x$ is bounded between $[-1,1]$, so we can assert that the value of $x^2\cos\frac1x$ is between $-x^2$ and $x^2$.  As $x$ approaches $0$, the values of the boundaries are both $0$, and thus the only possible value for $\lim_{x\to0}x^2\cos\frac1x$ is $0$.

anonymous · Answer

u need to show with a graph or/and showing as piecewise funtion

anonymous · Answer

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Haha, shoddy drawing aside, the parabolic functions are the boundaries, whereas the actual function is constrained by said boundaries. At $0$ there is only one possible region for the actual function to pass through,