Evaluate the limit: $ \lim_{x\to -1}\frac{x\ln(x+3) + \ln(2)} {x+1} $

Question

$$ \lim_{x\to -1}\frac{x\ln(x+3) + \ln(2)} {x+1} $$

I tried to separate the fraction and also a change of variable (x+3 = y+1) but I couldn't solve it. Maybe there's a trivial step that I'm just missing. Thanks.

Note: the person I was trying to help doesn't know how to use l'Hopital's rule.

Answer 1

Set $x+1=y$ to get

$$\lim_{y\to0}\dfrac{(y-1)\ln(y+2)+\ln2}y=\lim_{y\to0}\ln(y+2)-\lim_{y\to0}\dfrac{\ln(y+2)-\ln2}y$$

Now, $$\lim_{y\to0}\dfrac{\ln(y+2)-\ln2}y=\lim_{y\to0}\dfrac{\ln\left(1+\dfrac y2\right)}{\dfrac y2}\cdot\dfrac12=?$$

Answer 2

Let $f(x)=x\ln(x+3)$.

$$\lim_{x\to -1}\frac{x\ln(x+3) + \ln(2)} {x+1}=\lim_{x\to -1}\frac{f(x)-f(-1)}{x-(-1)}=f'(-1)$$