How to solve $\lim_{x\rightarrow2}\frac{(x-2)\ln(x-1)}{1-\cos(x-2)}$?

Question

We have the function $f(x)=\frac{(x-2)\ln(x-1)}{1-\cos(x-2)}$ and I'm supposed to check if I can set a value for $f(2)$ such that $f$ is continuous. So I'm trying to calculate the limit of $f$ as $x$ approaches $2$.

What I've done so far is the following $$\begin{align} \frac{(x-2)\ln(x-1)}{1-\cos(x-2)}=&\frac{(x-2)\ln(x-1)(1+\cos(x-2))}{(1-\cos(x-2))(1+\cos(x-2))}\\=&\frac{(x-2)\ln(x-1)(1+\cos(x-2))}{\sin^2(x-2)}\\=&\frac{(x-2)\ln(x-1)(1+\cos(x-2))(x-2)^2}{\sin^2(x-2)(x-2)^2}\\=&\frac{(x-2)\ln\frac{1}{1-x}(1+\cos(x-2))}{(x-2)(2-x)}\frac{2-x}{\sin(2-x)}\frac{2-x}{\sin(2-x)}\end{align}$$

Answer 1

If $\lim_\limits{x\to a} f(x)$ and $\lim_\limits{x\to a} g(x)$ exist and are bounded then $\lim_\limits{x\to a} f(x)g(x) = \left(\lim_\limits{x\to a} f(x)\right)\left(\lim_\limits{x\to a} g(x)\right)$

$\lim_\limits{x\to 2} \frac {(x-2)\ln(x-1)}{1-\cos(x-2)}\\ \lim_\limits{x\to 2} \frac {(x-2)\ln(x-1)(1+\cos(x-2)}{\sin^2(x-2)}\\ \lim_\limits{x\to 2} \frac {(x-2)^2\ln(x-1)(1+\cos(x-2)}{(x-2)\sin^2(x-2)}\\ \left(\lim_\limits{x\to 2} \frac {\ln(x-1)}{x-2}\right) \left(\lim_\limits{x\to 2}1+\cos(x-2)\right)\left( \lim_\limits{x\to 2}\frac{(x-2)^2}{\sin^2(x-2)}\right)$

$\left(\lim_\limits{x\to 2} \frac {\ln(x-1)}{x-2}\right) \left(2\right)\left(1\right)$

That leaves us with one remaining limit to evaluate.

$\lim_\limits{x\to 2} \frac {\ln(x-1)}{x-2}$.
Let $u = x-2$.

$\lim_\limits{u\to 0} \frac {\ln(1+u)}{u}\\ \lim_\limits{u\to 0} \frac 1u\ln(1+u)\\ \lim_\limits{u\to 0} \ln(1+u)^\frac 1u\\ \lim_\limits{u\to 0}(1+u)^\frac 1u = e \\ \lim_\limits{u\to 0} \ln(1+u)^\frac 1u = \ln e = 1$

Multiplying by the other factors gives

$\lim_\limits{x\to 2} \frac {(x-2)\ln(x-1)}{1-\cos(x-2)} = 2$

Answer 2

set $t=x-2$ then you have: \begin{align} \lim_{t \to 0}\frac{t\ln(t+1)}{1-\cos(t)}=&\lim_{t \to 0}\frac{t\ln(t+1)(1+\cos(t))}{(1-\cos(t))(1+\cos(t))}\\=&\lim_{t \to 0}\frac{t\ln(t+1)(1+\cos(t))}{\sin^2t}\\=&\lim_{t \to 0}\frac{t^2}{\sin^2(t)}\frac{\ln(t+1)}{t}(1+\cos(t))\\=&\lim_{t \to 0}(\frac{t}{\sin(t)})^2(t^{-1}\ln(t+1))(1+\cos(t)) \\=&\lim_{t \to 0}(\frac{t}{\sin(t)})^2\ln(t+1)^{\frac{1}{t}}(1+\cos(t))=1^2\ln(e)(1+1)=2 \end{align}

I made 2 assumptions first $\frac{t}{\sin(t)}$ goes to 1 and second $(t+1)^{\frac{1}{t}}$ goes to e. the proof for the first is here and the second can be considered the definition for e.

Answer 3

Let $v=x-2$. Then the limit we are trying to find is $$\lim_{v\rightarrow0}\frac{v\ln(1+v)}{1-\cos(v)}$$Multiply by $1+\cos(v)$ top and bottom: $$\lim_{v\rightarrow0}\frac{v\ln(1+v)+v\cos(v)\ln(1+v)}{\sin^2v}$$Because $\cos(v)$ approaches $1$ as $v$ approaches $0$, We could write the limit as $$2\lim_{v\rightarrow0}\frac{v\ln(1+v)}{\sin^2v}=2\lim_{v\rightarrow0}\frac{\ln(1+v)}{\sin v}$$ Substituting the series expansion of $\ln(1+v)$ when $|v|\le1$ and using that $$\lim_{v\rightarrow0}\frac v{\sin v}=1$$for each of the terms gives the answer.

Edit: Thanks to Ryszard Szwarc's comment, we could proceed as follows: $$\lim_{v\rightarrow0}\frac{\ln(1+v)}{\sin v}=\lim_{v\rightarrow0}\frac{\ln(1+v)}v\frac v{\sin v}=\lim_{v\rightarrow0}\frac{\ln(1+v)}v =\lim_{v\rightarrow0}\frac{\ln(1+v)-\ln(1)}v$$By the definition of the derivative: $$\lim_{v\rightarrow0}\frac{\ln(1+v)-\ln(1)}v=\frac d{dx}\ln(1+x)\bigg|_0=1$$

Answer 4

Indeed, if the limit as $x\to 2$ of $f$ exists, then you can extend the domain of definition for $f$ to include $2$ and make $f$ continuous at $2$ by defining $f(2)$ to be the value of the limit. Let $L:=\lim_{x\to 2}\tfrac{(x-2)\ln(x-1)}{1-\cos(x-2)}$.

By a simple change of variable ($x-2$ becomes $x$) we have $L=\lim_{x\to 0} \tfrac{x\ln(x+1)}{1-\cos x}$.

Now we can use two well known inequalities based on Taylor expansion, although these can be proved directly without knowing Taylor's theorem.

(*):$\tfrac{x}{x+1}\leq \ln(1+x)\leq x$ for all $x>-1$

(**): $\tfrac{x^2}{2}-\tfrac{x^3}{6}\leq 1-\cos x\leq \tfrac{x^2}{2}+\tfrac{x^3}{6}$ for all $x$.

Massaging these a bit, we get $\tfrac{1}{x+1}\leq \tfrac{\ln(1+x)}{x}\leq 1$ and $\tfrac{1}{2}-\tfrac{x}{6}\leq \tfrac{1-\cos x}{x^2}\leq \tfrac{1}{2}+\tfrac{x}{6}$.

This implies that $L_1:=\lim_{x\to 0} \tfrac{\ln(1+x)}{x} =1$ and $L_2:=\lim_{x\to 0} \tfrac{x^2}{1-\cos x}=2$.

We can recover $L=L_1\cdot L_2=2$.