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?

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 $\frac{121}{160}\approx 0.75625$.
