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Math  /  Calculus

Question18. [-/1 Points]
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SCALCET9M 4.4.058. ASK YOL
Find the limit. Use I'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. lim⁡x→0+(tan⁡(8x))x\lim _{x \rightarrow 0^{+}}(\tan (8 x))^{x} □\square Need Help? Read It

Studdy Solution

STEP 1

Assumptions
1. We need to find the limit: lim⁡x→0+(tan⁡(8x))x\lim _{x \rightarrow 0^{+}}(\tan (8 x))^{x}
2. The function is in the indeterminate form 000^0 as x→0+x \to 0^+.
3. We will use logarithmic properties and L'Hospital's Rule to solve this limit.

STEP 2

To handle the indeterminate form 000^0, take the natural logarithm of the function. Let y=(tan⁡(8x))xy = (\tan(8x))^x, then:
ln⁡y=xln⁡(tan⁡(8x))\ln y = x \ln(\tan(8x))

STEP 3

Now, find the limit of ln⁡y\ln y as x→0+x \to 0^+:
lim⁡x→0+ln⁡y=lim⁡x→0+xln⁡(tan⁡(8x))\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} x \ln(\tan(8x))

STEP 4

This limit is in the form 0⋅(−∞)0 \cdot (-\infty), which is indeterminate. Rewrite it as a fraction:
lim⁡x→0+xln⁡(tan⁡(8x))=lim⁡x→0+ln⁡(tan⁡(8x))1/x\lim_{x \to 0^+} x \ln(\tan(8x)) = \lim_{x \to 0^+} \frac{\ln(\tan(8x))}{1/x}

STEP 5

Now, the limit is in the form −∞∞\frac{-\infty}{\infty}, which is suitable for L'Hospital's Rule. Differentiate the numerator and the denominator:
Numerator: ddx[ln⁡(tan⁡(8x))]=1tan⁡(8x)⋅sec⁡2(8x)⋅8=8sec⁡2(8x)tan⁡(8x)\frac{d}{dx}[\ln(\tan(8x))] = \frac{1}{\tan(8x)} \cdot \sec^2(8x) \cdot 8 = \frac{8 \sec^2(8x)}{\tan(8x)}
Denominator: ddx[1/x]=−1x2\frac{d}{dx}[1/x] = -\frac{1}{x^2}

STEP 6

Apply L'Hospital's Rule:
lim⁡x→0+ln⁡(tan⁡(8x))1/x=lim⁡x→0+8sec⁡2(8x)tan⁡(8x)⋅(−x2)\lim_{x \to 0^+} \frac{\ln(\tan(8x))}{1/x} = \lim_{x \to 0^+} \frac{8 \sec^2(8x)}{\tan(8x)} \cdot (-x^2)

STEP 7

Simplify the expression:
=lim⁡x→0+−8x2⋅sec⁡2(8x)tan⁡(8x)= \lim_{x \to 0^+} -8x^2 \cdot \frac{\sec^2(8x)}{\tan(8x)}

STEP 8

As x→0+x \to 0^+, tan⁡(8x)≈8x\tan(8x) \approx 8x and sec⁡2(8x)≈1\sec^2(8x) \approx 1. Substitute these approximations:
=lim⁡x→0+−8x2⋅18x= \lim_{x \to 0^+} -8x^2 \cdot \frac{1}{8x}

STEP 9

Simplify further:
=lim⁡x→0+−x= \lim_{x \to 0^+} -x

STEP 10

Evaluate the limit:
=0= 0

STEP 11

Since lim⁡x→0+ln⁡y=0\lim_{x \to 0^+} \ln y = 0, we have ln⁡y=0\ln y = 0, which implies y=e0=1y = e^0 = 1.

STEP 12

Thus, the original limit is:
lim⁡x→0+(tan⁡(8x))x=1\lim_{x \to 0^+} (\tan(8x))^x = 1

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