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

Question

I have trouble calculating $\lim_{x \to 2} \left(\frac{1}{x-2}\cdot 2\ln(\frac{2}{x})\right)$. While I checked the result in Wolfram Alpha, it's $-1$, my result is $0$, please let me know what I'm doing wrong.

These are my calculations:

$$\lim_{x \to 2^+} \left(\frac{1}{x-2}\cdot 2\ln(\frac{2}{x})\right) = $$

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

$$=\frac{"-\infty"}{"\infty"}$$ so by L'hôpital rule:

$$\lim_{x \to 2^+} \left(\frac{\left(2\ln(\frac{2}{x})\right)'}{\left(\frac{1}{\frac{1}{x-2}}\right)'} \right) = $$

$$=\lim_{x \to 2^+} \left( \frac{-2\cdot\frac{1}{x}}{-\frac{1}{(x-2)^2}} \right) = \lim_{x \to 2^+} \frac{2(x-2)^2}{x} = 0$$

Answer 1

By applying L hospital rule , denominator becomes $$=\lim_{x \to 2^+} \left( \frac{-2\cdot\frac{1}{x}}{{1}} \right) =-1$$

You can write $$\left(\frac{1}{\frac{1}{x-2}}\right)'=(x-2)'$$ as $(x-2)'$ which is = 1

Answer 2

Write $$\frac{1}{x-2}\cdot 2\ln\left(\frac{2}{x}\right)= -2 \frac{\ln(x)-\ln(2)}{x-2}$$ and your limit is related to the derivative of $\ln$ at $x=2.$ $$\lim_{x \to 2} \left(\frac{1}{x-2}\cdot 2\ln\left(\frac{2}{x}\right)\right)=-2\ln'(x)|_{x=2}=-2\left(\frac{1}{x}\right)_{x=2} = -1$$

Answer 3

Write your limit in the form $$ \lim_{x\to 2}-2\frac{\ln x-\ln 2}{x-2}= -2\lim_{x\to 2}\frac{\ln x-\ln 2}{x-2} $$ Do you spot anything familiar?

Where is your error? The limit is not in the form $\infty/\infty$, but $0/0$. And the derivative of $$ \frac{1}{\frac{1}{x-2}} $$ is simply $1$.

Answer 4

I think you did some rather unnecessary steps.

Simplification is actually not too bad:

$$\lim_{x\rightarrow 2}\frac{2ln\frac{2}{x}}{x-2}$$

$$=\lim_{x\rightarrow 2}\frac{-2(\frac{x}{2})(\frac{2}{x^2})}{1}$$

$$=\lim_{x\rightarrow 2}(-\frac{2}{x})$$

$$=-1.$$

Answer 5

$$\lim_{x\to2^+}\left(\frac{2}{x-2}\ln[\frac{2}{x}]\right)=\lim_{x\to0^+}\left(\frac{2}{x}\ln\left[\frac{2+x-x}{x+2}\right]\right)=\lim_{x\to0^+}\left(\frac{2}{x}\ln\left[1-\frac{x}{x+2}\right]\right)\longrightarrow_0 \frac{2}{x}\left(-\frac{x}{x+2}\right)=-1$$

Answer 6

$$\left(\frac{1}{\frac{1}{x-2}}\right)' \neq -\frac{1}{(x-2)^2}$$ The derivative, in fact, is $1$.