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Question: The value of $\lim_{x \to (\frac{\pi}{10})^{+}} (\tan(\frac{3\pi}{10}+x))^{\tan 5x}$ is...

The value of lim⁡x→(π10)+(tan⁡(3π10+x))tan⁡5x\lim_{x \to (\frac{\pi}{10})^{+}} (\tan(\frac{3\pi}{10}+x))^{\tan 5x} is

A

1

B

0

C

e

D

1e\frac{1}{e}

Answer

1

Explanation

Solution

Let the limit be LL. L=lim⁡x→(π10)+(tan⁡(3π10+x))tan⁡5xL = \lim_{x \to (\frac{\pi}{10})^{+}} \left(\tan\left(\frac{3\pi}{10}+x\right)\right)^{\tan 5x} This is an indeterminate form of the type BEB^E. We use the property lim⁡x→af(x)g(x)=elim⁡x→ag(x)ln⁡f(x)\lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} g(x) \ln f(x)}. Let y=x−π10y = x - \frac{\pi}{10}. As x→(π10)+x \to (\frac{\pi}{10})^{+}, we have y→0+y \to 0^{+}. Then x=y+π10x = y + \frac{\pi}{10}. The exponent becomes tan⁡(5x)=tan⁡(5(y+π10))=tan⁡(π2+5y)=−cot⁡(5y)\tan(5x) = \tan\left(5\left(y+\frac{\pi}{10}\right)\right) = \tan\left(\frac{\pi}{2}+5y\right) = -\cot(5y). The base becomes tan⁡(3π10+x)=tan⁡(3π10+y+π10)=tan⁡(2π5+y)\tan\left(\frac{3\pi}{10}+x\right) = \tan\left(\frac{3\pi}{10}+y+\frac{\pi}{10}\right) = \tan\left(\frac{2\pi}{5}+y\right). So we need to evaluate the limit of the logarithm: ln⁡L=lim⁡y→0+tan⁡(5x)ln⁡(tan⁡(3π10+x))\ln L = \lim_{y \to 0^{+}} \tan(5x) \ln\left(\tan\left(\frac{3\pi}{10}+x\right)\right) ln⁡L=lim⁡y→0+(−cot⁡(5y))ln⁡(tan⁡(2π5+y))\ln L = \lim_{y \to 0^{+}} (-\cot(5y)) \ln\left(\tan\left(\frac{2\pi}{5}+y\right)\right) This is of the indeterminate form ∞⋅ln⁡(B)\infty \cdot \ln(B) where B=tan⁡(2π5)>1B = \tan(\frac{2\pi}{5}) > 1. We rewrite it as: ln⁡L=lim⁡y→0+−ln⁡(tan⁡(2π5+y))cot⁡(5y)\ln L = \lim_{y \to 0^{+}} \frac{-\ln\left(\tan\left(\frac{2\pi}{5}+y\right)\right)}{\cot(5y)} This is of the form −ln⁡(tan⁡(2π5))∞\frac{-\ln(\tan(\frac{2\pi}{5}))}{\infty}, which approaches 00. Alternatively, using L'Hopital's Rule: Let N(y)=−ln⁡(tan⁡(2π5+y))N(y) = -\ln\left(\tan\left(\frac{2\pi}{5}+y\right)\right) and D(y)=cot⁡(5y)D(y) = \cot(5y). N′(y)=−1tan⁡(2π5+y)⋅sec⁡2(2π5+y)=−sec⁡2(2π5+y)tan⁡(2π5+y)N'(y) = -\frac{1}{\tan(\frac{2\pi}{5}+y)} \cdot \sec^2\left(\frac{2\pi}{5}+y\right) = -\frac{\sec^2(\frac{2\pi}{5}+y)}{\tan(\frac{2\pi}{5}+y)}. D′(y)=−5csc⁡2(5y)D'(y) = -5\csc^2(5y). lim⁡y→0+N′(y)D′(y)=lim⁡y→0+−sec⁡2(2π5+y)tan⁡(2π5+y)−5csc⁡2(5y)=lim⁡y→0+sec⁡2(2π5+y)5tan⁡(2π5+y)csc⁡2(5y)\lim_{y \to 0^{+}} \frac{N'(y)}{D'(y)} = \lim_{y \to 0^{+}} \frac{-\frac{\sec^2(\frac{2\pi}{5}+y)}{\tan(\frac{2\pi}{5}+y)}}{-5\csc^2(5y)} = \lim_{y \to 0^{+}} \frac{\sec^2(\frac{2\pi}{5}+y)}{5\tan(\frac{2\pi}{5}+y)\csc^2(5y)} =lim⁡y→0+sec⁡2(2π5+y)5tan⁡(2π5+y)sin⁡2(5y)= \lim_{y \to 0^{+}} \frac{\sec^2(\frac{2\pi}{5}+y)}{5\tan(\frac{2\pi}{5}+y)} \sin^2(5y) As y→0+y \to 0^{+}, sin⁡(5y)≈5y\sin(5y) \approx 5y, so sin⁡2(5y)≈25y2\sin^2(5y) \approx 25y^2. The limit becomes: lim⁡y→0+sec⁡2(2π5)5tan⁡(2π5)(25y2)=0\lim_{y \to 0^{+}} \frac{\sec^2(\frac{2\pi}{5})}{5\tan(\frac{2\pi}{5})} (25y^2) = 0 So, ln⁡L=0\ln L = 0. Therefore, L=e0=1L = e^0 = 1.