## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{x\to1}\ln\Big(\frac{5-x^2}{1+x}\Big)=\ln2$
$$A=\lim\limits_{x\to1}\ln\Big(\frac{5-x^2}{1+x}\Big)$$ Let $f(x)=\ln x$ and $g(x)=\frac{5-x^2}{1+x}$, we have $$A=\lim\limits_{x\to1}f(g(x))$$ Since $\ln x$ is continuous on $(0,\infty)$, we can apply Theorem 8: $$A=f(\lim\limits_{x\to1}g(x))$$ $g(x)$ is a rational function, so it is continuous on $R$ except where $x=-1$. Therefore, $\lim\limits_{x\to1}g(x)=g(1)$ $A=f(g(1))$ $A=\ln\Big(\frac{5-1^2}{1+1}\Big)$ $A=\ln2$