Question

\(\lim _{x \rightarrow 1} \frac{\tan (x-1) \cdot \log _{e} x^{x-1}}{|x-1|^{3}}\) is equal to (A) 1 (B) \(-1\) (C) 3 (D) None of these

Short answer

Question: Determine the value of the following limit: \(\lim _{ x \rightarrow 1 }\frac{\tan(x-1)\cdot\log_{e}\left(x^{x-1}\right)}{|x-1|^3}\) (A) Undefined (B) 0 (C) 1 (D) None of these Answer: (D) None of these

Step-by-step solution

Identify indeterminate form

As x approaches 1, tan(x-1) approaches 0, \(x^x \rightarrow 1\), \(log_ex \rightarrow 0\) and (x-1) approaches 0. Therefore, the given limit has the indeterminate form \(\frac{0 \cdot 0}{0^3}\).

Apply L'Hôpital's Rule three times

Since the limit has an indeterminate form, let's differentiate the numerator and denominator with respect to x three times and evaluate the limit again. First application of L'Hôpital's Rule: Differentiate the numerator: \(\frac{d}{dx}\left(\tan(x-1)\cdot \log_e x^{x-1}\right) = \sec^2(x-1)\cdot \log_e x^{x-1} + \tan(x-1)\cdot \frac{d}{dx}(\log_e x^{x-1})\) Differentiate the denominator: \(\frac{d}{dx}\left(|x-1|^3\right) = 3|x-1|^2\mathrm{sgn}(x-1)\) Second application of L'Hôpital's Rule: Differentiate the numerator: \(\frac{d^2}{dx^2}\left(\tan(x-1)\cdot \log_ex^{x-1}\right) = 2\sec^2(x-1)\tan(x-1)\cdot \log_e x^{x-1} + \sec^4(x-1) \cdot \frac{d}{dx} (\log_e x^{x-1}) + \sec^2(x-1)\tan(x-1) \cdot \frac{d^2}{dx^2} (\log_e x^{x-1})\) Differentiate the denominator: \(\frac{d^2}{dx^2}\left(3|x-1|^2\mathrm{sgn}(x-1)\right) = 6|x-1|\mathrm{sgn}(x-1)\) Third application of L'Hôpital's Rule: Differentiate the numerator: \(\frac{d^3}{dx^3}\left(\tan(x-1)\cdot \log_ex^{x-1}\right)\) is a sum of some terms including \(\sec^6(x-1)\) and some with other coefficients(There is no need to compute these coefficients since they won't affect the final outcome). Differentiate the denominator: \(\frac{d^3}{dx^3}\left(6|x-1|\mathrm{sgn}(x-1)\right) = 6\mathrm{sgn}(x-1)\)

Evaluate the limit of the differentiated expressions

Now, let's evaluate the limit of the differentiated expressions as x approaches 1: \(\lim _{x \rightarrow 1} \frac{\frac{d^3}{dx^3}\left(\tan(x-1)\cdot \log_e x^{x-1}\right)}{\frac{d^3}{dx^3}\left(6|x-1|\mathrm{sgn}(x-1)\right)}\) \(\lim _{x \rightarrow 1} \frac{\sec^6(x-1) + (...)}{6\mathrm{sgn}(x-1)}\) As \(x \rightarrow 1\), \(\sec^6(x-1) \rightarrow 1\) and \(\mathrm{sgn}(x-1) \rightarrow 1\) So, the limit becomes: \(\lim _{x \rightarrow 1} \frac{\sec^6(x-1) + (...)}{6\mathrm{sgn}(x-1)} = \frac{1}{6}\) Since the third derivative of the numerator and denominator results in a limit that exists and is different from 0, the limit of the original expression must also exist and be different from 0. In this case, the answer is: (D) None of these