Question

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→1 3x x − 1 − 3 ln(x)

Answer 1

Answer:

3/2

Step-by-step explanation:

Given the limit of a function expressed as $\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})$, we are to evaluate it. To evaluate it, we will simply substitute x = 1 into the function since the variable x tends to 1.

$\lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= (\dfrac{3(1)}{1-1} - \dfrac{3}{ln1})\\\\= \dfrac{3}{0} - \dfrac{3}{0}\\\\= \infty - \infty (ind)$

Since we got an indeterminate function, we will find the LCM of the function and solve again.

$= \lim_{x \to 1} (\dfrac{3x}{x-1} - \dfrac{3}{lnx})\\\\= \lim_{x \to 1} \dfrac{3xlnx-3(x-1)}{(x-1)lnx}\\\\\\= \dfrac{3(1)ln(1)-3(1-1)}{(1-1)ln1}\\\\= \frac{3(0)-3(0)}{0(0)} \\\\= \frac{0}{0} (ind)$

Applying L'hospital rule;

$\frac{x}{y} = \lim_{x \to 1} \dfrac{d/dx(3xlnx-3(x-1))}{d/dx((x-1)lnx)}\\\\= \lim_{x \to 1} \dfrac{3x(\frac{1}{x})+ 3lnx-3)}{(x-1)\frac{1}{x} +lnx}\\\\= \lim_{x \to 1} \dfrac{3 + 3lnx-3}{(x-1)\frac{1}{x} +lnx}\\\\= \frac{3ln1}{(1-1)\frac{1}{1} +ln1}\\\\= \frac{0}{0} (ind)$

Applying L'hospital rule again;

$= \lim_{x \to 1} \dfrac{\frac{d}{dx} (3lnx)}{\frac{d}{dx} ((x-1)\frac{1}{x} +lnx)}\\\\= \lim_{x \to 1} \dfrac{\frac{3}{x} }{(x-1)\frac{-1}{x^2} + \frac{1}{x} +\frac{1}{x} }\\\\= \dfrac{\frac{3}{1} }{(1-1)\frac{-1}{1^2} + \frac{1}{1} +\frac{1}{1} }\\\\= \frac{3}{0(-1)+2}\\ \\= \frac{3}{2-0}\\ \\= 3/2$

Hence the limit of the function is 3/2.