Evaluating $\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$ and $\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$ I found these limits and I was unable to solve them due to the occurring indeterminations
$$\lim_{x \to 1}\frac{\sin(x^2-x)}{x^2-1}$$
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$
Can someone help me, please?
 A: Partial answer.
$\dfrac{x\sin (x(x-1))}{x(x-1)(x+1)}=$
$\dfrac{\sin (x(x-1))}{x(x-1)}\dfrac{x}{x+1}$;
Take the limit $x \rightarrow 1$.
Recall $\lim_{y \rightarrow 0}\dfrac{\sin y}{y}=1$.
A: Hint:
For the first,
$$\frac{\sin(x^2-x)}{x^2-1}=\frac{\sin (x^2-x)}{x^2-x}\frac{x^2-x}{x^2-1}=\frac{\sin (x^2-x)}{x^2-x} \frac{x(x-1)}{(x-1)(x+1)}$$
Don't forget, what is $\lim\limits_{t \to 0} \frac{\sin t}{t}$?
For the second , you can use l'Hopitals rule, or you might want to use the same trick as the first one, or maybe using the fact that $\lim\limits_{t \to 1} \frac{\ln t}{t-1}=?$
A: Since $\lim_{t \to 0}\frac{\log_{a}\left(t+1\right)}{t}=\log_{a}\left(e\right)$, 
we have:
$$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)}$$$$=\lim_{x \to 1}\frac{\ln\left(1+\left(\color{blue}{x^{2}+x-2}\right)\right)}{\color{blue}{x^{2}+x-2}}\cdot\frac{x^{2}+x-2}{x^{3}+x-2}\cdot\frac{\color{red}{x^{3}+x-2}}{\ln\left(1+\left(\color{red}{{x^{3}+x-2}}\right)\right)}=3/4$$
note that: $\frac{x^{2}+x-2}{x^{3}+x-2}=\frac{\left(x+2\right)\left(x-1\right)}{\left(x^{2}+x+1\right)\left(x-1\right)}$
A: For the second one $$\frac{\log(x^2+x-1)}{\log(x^3+x-1)}$$ using what Lord Shark the Unknown suggested, it becomes
$$\frac{\log \left(1+3t+t^2\right)}{\log \left(1+4t+3 t^2+t^3\right)}$$ Use the Taylor expansion of $\log(1+\epsilon)$ with $\epsilon=(3t+t^2)$ for the numerator and $\epsilon=(4t+3 t^2+t^3)$ for the denominator and use the binomial expansion or Taylor series to get 
$$\frac{\log \left(1+3t+t^2\right)}{\log \left(1+4t+3 t^2+t^3\right)}=\frac{3 t-\frac{7 }{2}t^2+O\left(t^3\right) } {4 t-5 t^2+O\left(t^3\right) }$$ Now, long division to get 
$$\frac{\log \left(1+3t+t^2\right)}{\log \left(1+4t+3 t^2+t^3\right)}=\frac{3}{4}+\frac{t}{16}+O\left(t^2\right)$$ which shows the limit and also how it is approached.
Just by curiosity, use your pocket calculator with $t=\frac 1 {10}$ (quite far away from $0$). You should get
$$\frac{\log \left(\frac{131}{100}\right)}{\log \left(\frac{1431}{1000}\right)}\approx 0.75348$$ while the above truncated expansion gives $\frac{121}{160}\approx 0.75625$.
A: $$\lim_{x \to 1}\frac{\ln(x^2+x-1)}{\ln(x^3+x-1)} = \lim_{x \to 1} \frac{(2x+1)\cdot(x^3+x-1)}{(3x^2+1)\cdot(x^2+x-1)} = \frac{3}{4}$$ (using L'Hospital)
