Question

$\lim _{x \rightarrow 1} \frac{(\ell n(1+x)-\ell \operatorname{n} 2)\left(3.4^{x-1}-3 x\right)}{\left[(7+x)^{1 / 3}-(1+3 x)^{1 / 2}\right] \cdot \sin (x-1)}$ equals (A) \(\frac{9}{4}\) en \(\frac{4}{\mathrm{e}}\) (B) \(\frac{9}{4}\) en \(\frac{\mathrm{e}}{4}\) (C) \(\frac{4}{9} \ell \mathrm{n} \frac{\mathrm{e}}{4}\) (D) None of these

Short answer

Question: Determine the limit of the expression as \(x\) approaches 1: \(\lim_{x\rightarrow 1}\frac{(\ln(1+x)-\ln2)(3.4^{x-1}-3x)}{[(7+x)^{\frac{1}{3}}-(1+3x)^{\frac{1}{2}}]\cdot\sin(x-1)}\) Answer: (D) None of these

Step-by-step solution

Determine if L'Hopital's Rule is needed

First, evaluate the expression when \(x=1\) and check if it's in the \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) indeterminate form. Plugging in \(x=1\), we get: \(\lim_{x\rightarrow 1}\frac{(\ln(1+1)-\ln2)(3.4^{1-1}-3\cdot 1)}{[(7+1)^{\frac{1}{3}}-(1+3\cdot 1)^{\frac{1}{2}}]\cdot\sin(1-1)}\) Simplifying the numerator and denominator: \(\lim_{x\rightarrow 1}\frac{(\ln2-\ln2)(1-3)}{[2^{\frac{1}{3}}-2^{\frac{1}{2}}]\cdot0}\) The expression is in the \(\frac{0}{0}\) indeterminate form, so we can use L'Hopital's Rule.

Apply L'Hopital's Rule

Using L'Hopital's Rule, we take the derivative of the numerator and the denominator with respect to \(x\): The derivative of the numerator with respect to \(x\): \(\frac{d}{dx}\Big[(\ln(1+x)-\ln 2)(3.4^{x-1}-3x)\Big]\) We apply the product rule, where \(u = (\ln(1+x)-\ln 2)\) and \(v = (3.4^{x-1}-3x)\): \(u' = \frac{1}{1+x}\) and \(v' = 3.4^{x-1}\ln(3.4)-3\) So, the derivative of the numerator is: \(u'v + uv' = \frac{1}{1+x}(3.4^{x-1}-3x) + (\ln(1+x)-\ln 2)(3.4^{x-1}\ln(3.4)-3)\) The derivative of the denominator with respect to \(x\): \(\frac{d}{dx}\Bigg[\Big[(7+x)^{\frac{1}{3}}-(1+3x)^{\frac{1}{2}}\Big] \cdot\sin(x-1)\Bigg]\) We apply the product rule once again, where \(p = [(7+x)^{\frac{1}{3}}-(1+3x)^{\frac{1}{2}}]\) and \(q = \sin(x-1)\): \(p' = \frac{1}{3}(7+x)^{-\frac{2}{3}} - \frac{3}{2}(1+3x)^{-\frac{1}{2}}(3)\) and \(q' = \cos(x-1)\) So, the derivative of the denominator is: \(p'q + pq' = \Big(\frac{1}{3}(7+x)^{-\frac{2}{3}} - \frac{9}{2}(1+3x)^{-\frac{1}{2}}\Big)\sin(x-1) + \\ [(7+x)^{\frac{1}{3}}-(1+3x)^{\frac{1}{2}}]\cos(x-1)\) Now the limit becomes: $\lim_{x\rightarrow 1} \frac{\frac{1}{1+x}(3.4^{x-1}-3x) + (\ln(1+x)-\ln 2)(3.4^{x-1}\ln(3.4)-3)} {\Big(\frac{1}{3}(7+x)^{-\frac{2}{3}} - \frac{9}{2}(1+3x)^{-\frac{1}{2}}\Big)\sin(x-1) + [(7+x)^{\frac{1}{3}}-(1+3x)^{\frac{1}{2}}]\cos(x-1)}$ Evaluate the new limit as \(x\rightarrow 1\): \(\frac{\frac{1}{2}(1-3) + (0)(0-3)}{\Big(\frac{1}{3}(8)^{-\frac{2}{3}} - \frac{9}{2}(4)^{-\frac{1}{2}}\Big)\sin(0) + [(8)^{\frac{1}{3}}-(4)^{\frac{1}{2}}]\cos(0)}\) Simplify and notice that the sine function term goes to zero: \(\frac{-\frac{1}{2}}{2^{\frac{1}{3}} -2^{\frac{1}{2}}}\) Now we compare our answer to the given options: (A) \(\frac{9}{4}\) en \(\frac{4}{\mathrm{e}}\) (B) \(\frac{9}{4}\) en \(\frac{\mathrm{e}}{4}\) (C) \(\frac{4}{9} \ell \mathrm{n} \frac{\mathrm{e}}{4}\) (D) None of these Our answer does not match any of the given options, so the correct answer is: (D) None of these