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Math  /  Calculus

QuestionFind the follow
2. lim⁡x→0x21−cos⁡(3x)\lim _{x \rightarrow 0} \frac{x^{2}}{1-\cos (3 x)}

Studdy Solution

STEP 1

What is this asking? We need to find the limit of a fraction involving x2x^2 and 1−cos⁡(3x)1 - \cos(3x) as xx approaches 0. Watch out! Direct substitution gives us 00\frac{0}{0}, which is an indeterminate form.
We'll need a clever trick to evaluate this limit!

STEP 2

1. Rewrite using a trig identity
2. Simplify the expression
3. Evaluate the limit

STEP 3

Let's recall the half-angle identity: cos⁡(2θ)=1−2sin⁡2(θ) \cos(2\theta) = 1 - 2\sin^2(\theta) We can rewrite this as: 2sin⁡2(θ)=1−cos⁡(2θ) 2\sin^2(\theta) = 1 - \cos(2\theta) We want something that looks like 1−cos⁡(3x)1 - \cos(3x), so let's set 2θ=3x2\theta = 3x, which means θ=3x2\theta = \frac{3x}{2}.
Substituting, we get: 2sin⁡2(3x2)=1−cos⁡(3x) 2\sin^2\left(\frac{3x}{2}\right) = 1 - \cos(3x) Now we have a shiny new way to express 1−cos⁡(3x)1 - \cos(3x)!

STEP 4

Let's substitute this into our original expression: lim⁡x→0x21−cos⁡(3x)=lim⁡x→0x22sin⁡2(3x2) \lim_{x \rightarrow 0} \frac{x^2}{1 - \cos(3x)} = \lim_{x \rightarrow 0} \frac{x^2}{2\sin^2\left(\frac{3x}{2}\right)} Much better!

STEP 5

We know that lim⁡u→0sin⁡(u)u=1\lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 1.
We want to use this, so let's rewrite our expression: lim⁡x→0x22sin⁡2(3x2)=lim⁡x→012⋅x2sin⁡2(3x2) \lim_{x \rightarrow 0} \frac{x^2}{2\sin^2\left(\frac{3x}{2}\right)} = \lim_{x \rightarrow 0} \frac{1}{2} \cdot \frac{x^2}{\sin^2\left(\frac{3x}{2}\right)} We can further rewrite this as: lim⁡x→012⋅(xsin⁡(3x2))2 \lim_{x \rightarrow 0} \frac{1}{2} \cdot \left(\frac{x}{\sin\left(\frac{3x}{2}\right)}\right)^2

STEP 6

To make it look even *more* like our limit identity, let's multiply by (3/2)2(3/2)2\frac{(3/2)^2}{(3/2)^2} (which is the same as multiplying by one!): lim⁡x→012⋅(32x32sin⁡(3x2))2=lim⁡x→012⋅(23)2⋅(3x2sin⁡(3x2))2 \lim_{x \rightarrow 0} \frac{1}{2} \cdot \left(\frac{\frac{3}{2}x}{\frac{3}{2}\sin\left(\frac{3x}{2}\right)}\right)^2 = \lim_{x \rightarrow 0} \frac{1}{2} \cdot \left(\frac{2}{3}\right)^2 \cdot \left(\frac{\frac{3x}{2}}{\sin\left(\frac{3x}{2}\right)}\right)^2

STEP 7

As xx approaches 0, 3x2\frac{3x}{2} also approaches 0.
Therefore, we can use our limit identity: lim⁡x→0sin⁡(3x2)3x2=1 \lim_{x \rightarrow 0} \frac{\sin\left(\frac{3x}{2}\right)}{\frac{3x}{2}} = 1 So, lim⁡x→03x2sin⁡(3x2)=11=1 \lim_{x \rightarrow 0} \frac{\frac{3x}{2}}{\sin\left(\frac{3x}{2}\right)} = \frac{1}{1} = 1

STEP 8

Substituting this back into our expression, we get: lim⁡x→012⋅(23)2⋅(3x2sin⁡(3x2))2=12⋅49⋅(1)2=29 \lim_{x \rightarrow 0} \frac{1}{2} \cdot \left(\frac{2}{3}\right)^2 \cdot \left(\frac{\frac{3x}{2}}{\sin\left(\frac{3x}{2}\right)}\right)^2 = \frac{1}{2} \cdot \frac{4}{9} \cdot (1)^2 = \frac{2}{9}

STEP 9

The limit is 29\frac{2}{9}.

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