Solve $\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1}$ without using L'Hôpital's

Question

I tried:

$$\lim_{x \rightarrow 1}\frac{3\ln(x) - x^2+1}{x-1} = \\ \frac{3\ln(x) + (1-x^2)}{-1(1-x)} = \\ \frac{3\ln(x)+ (1-x)(1+x)}{-(1-x)} = \\ \frac{3\ln(x)}{x-1} + \frac{1+x}{-1} = \\ \frac{\ln{x^3}}{x-1} - 1-x = \\ ???$$

What do I do next? Remember, I can't use L'Hôpital.

Answer 1

Substitute $$y=x-1$$ to get $$\frac{\ln(x^3)}{x-1}=\frac{3\ln(y+1)}{y}$$

and now use $$\ln(y+1)=y+O(y^2)$$

Answer 2

Recall that

$$\ln(x^3)=3\ln(x)$$

and

$$\lim_{x\to1}\frac{\ln(x)}{x-1}=1$$

which should give you

$$\lim_{x\to1}\frac{3\ln(x)}{x-1}-1-x=3-1-1=\boxed1$$

Answer 3

$$\lim_{x\to1}\frac{\ln x}{x-1}=1$$ is a basic limit from high school, and expresses the logarithm is differentiable at $x=1$ and has derivative equal to $1$.