A calculus problem by Sabhrant Sachan

Express the given limit as $\lim_{x\rightarrow \infty} \dfrac{1}{x\left(\ln(x)\right)^3} + \lim_{x\rightarrow \infty} \dfrac{1}{\left(\ln(x)\right)^3} + \lim_{x\rightarrow \infty} \dfrac{x}{\left(\ln(x)\right)^3}$

Since $\lim_{x\rightarrow \infty} x\left(\ln(x)\right)^3 = \infty$ and $\lim_{x\rightarrow \infty} \left(\ln(x)\right)^3 = \infty$ , $\lim_{x\rightarrow \infty} \dfrac{1}{x\left(\ln(x)\right)^3} = 0$ and $\lim_{x\rightarrow \infty} \dfrac{1}{\left(\ln(x)\right)^3} = 0$ We then evaluate $\lim_{x\rightarrow \infty} \dfrac{x}{\left(\ln(x)\right)^3}$ There are few ways to evaluate the limit.

L'Hôpital's Rule

Since $\lim_{x\rightarrow \infty} x = \infty$ and $\lim_{x\rightarrow \infty} \left(\ln(x)\right)^3 = \infty$ , which leads to indeterminate form $\infty / \infty$ , by L'Hôpital's Rule , $\begin{array}{rl} \lim_{x\rightarrow \infty} \dfrac{\frac{d}{dx}(x)}{\frac{d}{dx}\left(\left(\ln(x)\right)^3\right)} &= \lim_{x\rightarrow \infty} \dfrac{1}{\frac{3}{x}\ln^2(x)}\\ &= \lim_{x\rightarrow \infty} \dfrac{x}{3\ln^2(x)}\\ &= \dfrac{1}{3} \lim_{x\rightarrow \infty} \dfrac{x}{\ln^2(x)} \end{array}$ By the same reason, $\begin{array}{rl} \dfrac{1}{3} \lim_{x \rightarrow \infty} \dfrac{x}{\ln^2(x)} &= \dfrac{1}{3} \lim_{x\rightarrow \infty} \dfrac{1}{\frac{2}{x}\ln(x)}\\ &= \dfrac{1}{6} \lim_{x \rightarrow \infty} \dfrac{x}{\ln(x)}\\ &= \dfrac{1}{6} \lim_{x \rightarrow \infty} \dfrac{1}{\frac{1}{x}}\\ &= \dfrac{1}{6} \lim_{x \rightarrow \infty} x\\ &= \infty \end{array}$

Logarithm Trick

Instead of L'Hôpital's Rule, we can evaluate the limit by $\ln$ i.e. $\begin{array}{rl} y &= \lim_{x\rightarrow \infty} \dfrac{x}{\ln^3(x)}\\ \ln(y) &= \lim_{x \rightarrow \infty} \ln\left(\dfrac{x}{\ln^3(x)}\right)\\ &= \lim_{x \rightarrow \infty} \ln(x) - 3 \lim_{x \rightarrow \infty} \ln\left( \ln(x)\right)\\ &= \lim_{x \rightarrow \infty} \ln(x) + 0\\ &= \lim_{x \rightarrow \infty} \ln(x) \end{array}$ Since $\lim_{x \rightarrow \infty} \ln(x) = \infty$ , $y = \infty$ , which proves that $\boxed{\text{limit does not exist}}$ .

I answered 0 as the problem told me lol...