Computing $\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$ 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
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2} \implies 0$$
 A: 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 $\lim_{x\to0}\frac{e^x-1}{x}=1$.
A: 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.
A: $\log(x+y)=\log(x)+\log(y)$ can't be possible.
$$\lim_{x \to 1}\frac{\log(x)}{x^2+x-2}$$
$$=\lim_{x \to 1} \frac{1}{2x^{2}+x} (\text{using L'hospital })$$
$$=\frac{1}{3}$$
A: \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!
A: $\dfrac{1}{x+2}\dfrac {\log x} {x-1};$
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?
A: 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}
