CalculusMediumLimits and Continuity ShareLimits and Continuity · Medium problemUse the Squeeze Theorem to evaluate limx→0x2cos (1x)\displaystyle\lim_{x\to 0}x^2\cos\!\left(\frac{1}{x}\right)x→0limx2cos(x1). State the bounding functions.A−x3≤x2cos(1/x)≤x2-x^{3}\le x^2\cos(1/x)\le x^2−x3≤x2cos(1/x)≤x2B−x4≤x2cos(1/x)≤x2-x^4\le x^2\cos(1/x)\le x^2−x4≤x2cos(1/x)≤x2C−x2≤x2cos(1/x)≤x3-x^2\le x^2\cos(1/x)\le x^3−x2≤x2cos(1/x)≤x3D−x2≤x2cos(1/x)≤x2-x^2\le x^2\cos(1/x)\le x^2−x2≤x2cos(1/x)≤x2Check Show answer Walk through it step by step Practice more like this Review this concept