Computing $\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$

Question

I'm trying to compute the following limit and would greatly appreciate your feedback:

$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$$

Given that $\log(x)$ can be replaced with $\log(x+1)$ because of the identity

$$\log(x) + \log(1) = \log(x) + 0$$

Then using the series:

$$\log(x+1) = x + o(x^n)$$

Then dividing by $x^2+x-2$, I get:

$$\frac{x}{x^2+x-2} = \frac{x + o(x^n)}{(x-1)(x+2)}$$

when $x = 1$

$$\frac{x + o(x^n)}{(x-1)(x+2)} = \frac{1}{0} = 0$$

hence

$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2} \implies 0$$

Why do you get to replace $\log x$ with a series for $\log(x+1)$? — Ninad Munshi, Apr 01 '21 at 16:00
You can't replace $A$ by $B$ unless $A=B$. Further we don't have $\log x=x+o(x^n) $ as $x\to 1$. — Paramanand Singh, Apr 01 '21 at 16:05
Are you aware of the limit formula $\lim_{x\to 1}\frac {\log x} {x-1}=1$? Use it here to get the desired answer. — Paramanand Singh, Apr 01 '21 at 16:06
L'Hospital rule may be of help here: $\frac{\log x}{x^2+x-2}\sim \frac{1}{x(2x+1)}\xrightarrow{x\rightarrow1}\frac13$ — Mittens, Apr 01 '21 at 16:07

score 1 · Accepted Answer · answered Apr 01 '21 at 16:05

1

Let $x = e^t$. Then $t\to 0$ as $x\to 1$. The limit then becomes $$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2} = \lim_{t\to0}\frac{t}{e^{2t}+e^t-2} = \lim_{t\to0}\color{red}{\frac{t}{e^t-1}}\frac{1}{e^t+2} = \color{red}1\cdot \frac{1}{1+2} = \frac{1}{3}$$ $\color{red}{\text{Using}}$ the well-known limit $\lim_{x\to0}\frac{e^x-1}{x}=1$.

answered Apr 01 '21 at 16:05

VIVID

11,667

The known limit really helps me understand this better! Thank you – SSDean Apr 01 '21 at 16:12
@SSDean Glad that it helps. You can accept the answer if it solves your issue :) – VIVID Apr 01 '21 at 16:14

score 1 · Answer 2 · answered Apr 01 '21 at 16:23

1

If you love Squeeze, you may want to try:

$$\frac{\frac1x (x-1)}{(x+2)(x-1)}≤\frac{\ln x}{x^2+x-2}≤\frac{x-1}{(x-1)(x+2)}$$

Then, apply the squeeze theorem.

answered Apr 01 '21 at 16:23

lone student

1

score 0 · Answer 3 · answered Apr 01 '21 at 16:07

0

$\log(x+y)=\log(x)+\log(y)$ can't be possible.

$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$$

$$=\lim_{x \to 1} \frac{1}{2x^{2}+x} (\text{using L'hospital })$$

$$=\frac{1}{3}$$

answered Apr 01 '21 at 16:07

A learner

2,871

score 0 · Answer 4 · answered Apr 01 '21 at 16:14

\begin{gather*} \lim _{x\rightarrow 1^{+}}\frac{\ln x}{( x-1)( x+2)} =\lim _{h\rightarrow 0}\frac{\ln( 1+h)}{h( h+3)}\\ =\lim _{h\rightarrow 0}\frac{\ln( 1+h)}{h} \cdotp \lim _{h\rightarrow 0}\frac{1}{h+3} =\frac{1}{3}\\ \lim _{x\rightarrow 1^{-}}\frac{\ln x}{( x-1)( x+2)} =\lim _{h\rightarrow 0}\frac{\ln( 1-h)}{-h( 3-h)} =\lim _{h\rightarrow 0}\frac{\ln( 1-h)}{h( h-3)}\\ =\lim _{h\rightarrow 0}\frac{\ln( 1-h)}{h} \cdotp \lim _{h\rightarrow 0}\frac{1}{h-3} =\frac{1}{3}\\ \end{gather*} Hope this helps!

Peter Szilas · Answer 5 · 2021-04-01T16:25:52.497

0

$\dfrac{1}{x+2}\dfrac {\log x} {x-1};$

Let |(x-1)|<<1, i.e. be small;

$\log (1+(x-1))=(x-1) - $

$(x-1)^2/2+(x-1)^3/3+..;$

Can you finish?

edited Apr 01 '21 at 16:25

answered Apr 01 '21 at 16:18

Peter Szilas

21,123

score 0 · Answer 6 · answered Apr 01 '21 at 16:18

0

You can do it as follows:\begin{align}\lim_{x\to1}\frac{\log x}{x^2+x-2}&=\lim_{x\to1}\frac{\log x}{(x-1)(x+2)}\\&=\lim_{x\to1}\frac{\log(x)-\log(1)}{x-1}\times\lim_{x\to1}\frac1{x+2}\\&=\log'(1)\times\frac13\\&=\frac13.\end{align}

answered Apr 01 '21 at 16:18

José Carlos Santos

440,053

Computing $\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$

6 Answers6