Question

Solve the limit without using L'Hôpital's rule

$\lim \limits_{ x \to \infty} \left[ k \ln(x) - x \right]$
... where k is a constant value.​

Answer 1

Given,

$\small\displaystyle\longrightarrow L=\lim_{x\to\infty}[k\ln x-x]$

On putting x = ∞ directly we get the indeterminate form ±∞ - ∞. But if k = 0 then L = -∞ as the function reduces to -x.

Taking x common in RHS,

$\small\displaystyle\longrightarrow L=\lim_{x\to\infty}x\left(\dfrac{k\ln x}{x}-1\right)$

$\small\text{ \displaystyle\longrightarrow L=\lim_{x\to\infty}\dfrac{\left(\dfrac{k\ln x}{x}-1\right)}{\left(\dfrac{1}{x}\right)} }$

$\small\text{ \longrightarrow L=\dfrac{\displaystyle\lim_{x\to\infty}\left(\dfrac{k\ln x}{x}-1\right)}{\displaystyle\lim_{x\to\infty}\left(\dfrac{1}{x}\right)} }$

$\small\text{ \longrightarrow L=\dfrac{\displaystyle k\cdot\lim_{x\to\infty}\left(\dfrac{\ln x}{x}\right)-1}{\displaystyle\lim_{x\to\infty}\left(\dfrac{1}{x}\right)}\quad\dots(1) }$

Let us evaluate the limit,

$\small\displaystyle\longrightarrow L_1=\lim_{x\to\infty}\dfrac{\ln x}{x}$

Let $\smallu=\dfrac{1}{x}$ so that $\smallu\to0$ when $\smallx\to\infty.$ Then,

$\small\text{ \displaystyle\longrightarrow L_1=\lim_{u\to0}\dfrac{\ln\left(\dfrac{1}{u}\right)}{\left(\dfrac{1}{u}\right)} }$

$\small\displaystyle\longrightarrow L_1=-\lim_{u\to0}(u\ln u)$

Now we need to consider the Taylor expansion,

$\small\longrightarrow\ln u=(u-1)-\dfrac{(u-1)^2}{2}+\dfrac{(u-1)^3}{3}-\dfrac{(u-1)^4}{4}+\,\dots$

So,

$\small\displaystyle\longrightarrow L_1=-\lim_{u\to0}\left(u\left((u-1)-\dfrac{(u-1)^2}{2}+\dfrac{(u-1)^3}{3}-\dfrac{(u-1)^4}{4}+\,\dots\right)\right)$

$\small\displaystyle\longrightarrow L_1=-\lim_{u\to0}\left(u(u-1)-\dfrac{u(u-1)^2}{2}+\dfrac{u(u-1)^3}{3}-\dfrac{u(u-1)^4}{4}+\,\dots\right)$

$\small\displaystyle\longrightarrow L_1=0$

Then (1) becomes,

$\small\text{ \longrightarrow L=\dfrac{k(0)-1}{\left(\dfrac{1}{\infty}\right)} }$

$\small\longrightarrow L=\dfrac{-1}{0}$

$\small\text{ \longrightarrow\underline{\underline{L=-\infty}} }$