Finding $\lim_\limits{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$ analytically

Question

$$\lim_{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$$

I can solve this easily by generating the graph with my calculator, but is there is a way to do this analytically?

Answer 1

Since the limit of both the top and bottom is $\infty$ alone, l'hopital's rule gives us \begin{align*} \lim_{x \to \infty}\frac{\log_{1.1} x}{x} &= \lim_{x \to \infty}\frac{\ln x}{(\ln 1.1) x} \\ &= \lim_{x \to \infty}\frac{(1/x)}{(\ln 1.1)} \\ &= 0. \end{align*}

Answer 2

hint: L'hospitale's rule ! . Have you learned this formula yet?

Answer 3

Let $g(x)=\ln(x)-\sqrt{x}$

we have $g(1)=0$ and

$$g'(x)=\frac{1}{2x}(2-\sqrt{x})$$

$g$ is decreasing at $[4,+\infty)$

thus

$$\forall x\geq 4 \;\;0<\frac{\log_{1.1}(x)}{x}\leq \frac{1}{\ln(1.1)\sqrt{x}}$$

and squeeze theorem.

Answer 4

Use the definition: $\;\log_{1.1}x=\dfrac{\ln x}{\ln 1.1}$, so $$\frac{\log_{1.1}x}{x}=\frac 1{\ln 1.1}\dfrac{\ln x}{x}\xrightarrow[x\to+\infty]{}\frac 1{\ln 1.1}\cdot 0=0.$$

Answer 5

I thought it might be instructive to present an approach that relies on elementary tools only. To that end, we begin with the following primer.

PRIMER: BOUNDS FOR THE LOGARITHM FUNCTION

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{x-1}\le \log(x)\le x-1} \tag 1$$

for $x<1$.

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Now, we note that $\log_{1.1}(x)=\frac{\log(x)}{\log(1.1)}$. In addition, we note that for any number $a$, we have $\log(x^a)=a\log(x)$.

Using $(1)$, we have for $a>0$,

$$\frac{x^{-1}-x^{-(a+1)}}{a}\le \frac{\log(x)}{x}\le \frac{x^{a-1}-x^{-1}}{a} \tag 2$$

Surely, $(2)$ is valid for $0<a<1$. Then, applying the squeeze theorem to $(2)$ with $0<a<1$ (e.g., $a=1/2$) yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\frac{\log_{1.1}(x)}{x}=0}$$

and we are done!