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Applying the squeeze theorem to (1-cos x)/x^2

We want to evaluate lim⁡x→01−cos⁡(x)x2\lim_{x \to 0} \frac{1-\cos(x)}{x^2} using the Squeeze Theorem, building on the known fact that lim⁡x→01−cos⁡(x)x=0\lim_{x\to 0} \frac{1-\cos(x)}{x} = 0 and lim⁡x→0sin⁡(x)x=1\lim_{x \to 0} \frac{\sin(x)}{x} = 1.

Recall the identity 1−cos⁡(x)x2=1−cos⁡(x)x⋅1x=(1−cos⁡(x)x)⋅(sin⁡(x)x)⋅11+cos⁡(x)⋅(1+cos⁡(x))\frac{1-\cos(x)}{x^2} = \frac{1-\cos(x)}{x}\cdot\frac{1}{x} = \left(\frac{1-\cos(x)}{x}\right)\cdot\left(\frac{\sin(x)}{x}\right)\cdot\frac{1}{1+\cos(x)}\cdot(1+\cos(x))

Instead of chasing that route, use the standard approach: multiply 1−cos⁡(x)x2\dfrac{1-\cos(x)}{x^2} by 1+cos⁡(x)1+cos⁡(x)\dfrac{1+\cos(x)}{1+\cos(x)} to rewrite it in terms of sin⁡(x)\sin(x), and then use the known limit laws for sin⁡(x)x\dfrac{\sin(x)}{x} together with the Squeeze Theorem (via the fact that cos⁡(x)\cos(x) is squeezed between values approaching 11) to evaluate

lim⁡x→01−cos⁡(x)x2.\lim_{x \to 0} \frac{1-\cos(x)}{x^2}.

What is the value of this limit?

Hint

Multiply numerator and denominator by 1+cos⁡(x)1+\cos(x) to turn 1−cos⁡(x)1-\cos(x) into sin⁡2(x)\sin^2(x) using the Pythagorean identity. Then split the resulting expression into a product of (sin⁡xx)2\left(\frac{\sin x}{x}\right)^2 and 11+cos⁡x\frac{1}{1+\cos x}, and evaluate each factor's limit as x→0x \to 0, using that cos⁡(x)\cos(x) is squeezed toward 11.

Step-by-step solution

Multiply the expression by 1+cos⁡(x)1+cos⁡(x)\frac{1+\cos(x)}{1+\cos(x)} to get 1−cos⁡(x)x2=(1−cos⁡(x))(1+cos⁡(x))x2(1+cos⁡(x))=1−cos⁡2(x)x2(1+cos⁡(x))=sin⁡2(x)x2(1+cos⁡(x)).\frac{1-\cos(x)}{x^2}=\frac{(1-\cos(x))(1+\cos(x))}{x^2(1+\cos(x))}=\frac{1-\cos^2(x)}{x^2(1+\cos(x))}=\frac{\sin^2(x)}{x^2(1+\cos(x))}. This rewrites the original 0/0 form into a product of factors whose limits are already known.

Now write this as (sin⁡(x)x)2⋅11+cos⁡(x).\left(\frac{\sin(x)}{x}\right)^2 \cdot \frac{1}{1+\cos(x)}. Since lim⁡x→0sin⁡(x)x=1\lim_{x\to 0}\frac{\sin(x)}{x}=1 (a Squeeze Theorem result) and cos⁡(x)\cos(x) is squeezed near 11 as x→0x\to 0 so 1+cos⁡(x)→21+\cos(x)\to 2, the product rule for limits gives lim⁡x→01−cos⁡(x)x2=(1)2⋅12=12.\lim_{x\to 0}\frac{1-\cos(x)}{x^2} = (1)^2 \cdot \frac{1}{2} = \frac{1}{2}.

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