Limit of x^2 sin(1/x) as x approaches 0

The limit of x^2 sin(1/x) as x approaches 0 is equal to 0, and it is denoted by lim_x→0 x² sin(1/x) = 0. So the limit formula of x² sin(1/x) when x tends to zero is given by

lim_x→0 x² sin(1/x) = 0.

We will now find the limit of x² sin(1/x) using the Sandwich/Squeeze theorem of limits.

Lim_x→0 x² sin(1/x) using Squeeze Theorem

We know that sinθ lies between -1 and 1 for any real values of θ. That is,

-1 ≤ sinθ ≤ 1.

So we obtain that

-1 ≤ sin(1/x) ≤ 1.

As x² ≥ 0, it follows that

-x² ≤ x² sin(1/x) ≤ x².

Now taking the limit lim_x→0 on both sides, we obtain that

lim_x→0 (-x²) ≤ lim_x→0 x² sin(1/x) ≤ lim_x→0 x².

⇒ 0 ≤ lim_x→0 x² sin(1/x) ≤ 0.

So by the Squeeze theorem, we get that lim_x→0 x² sin(1/x) = 0.

Thus, the limit of x² sin(1/x) is equal to 0 when x tends to 0, and this is proved using the Squeeze theorem. That is,

limx→0 x2 sin(1/x) = 0

More Limits:

Limit of (1+ 1/x)^x when x→∞

Limit of cos(1/x) when x→∞

Limit of cos(1/x) when x→0

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