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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.

3 Answers 3

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

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

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    Note that $y$ tends to $0$, if $x$ tends to $1$. – 2017-02-18
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Recall that

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

and

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

which should give you

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

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    My book mentions $\lim_{x\to0}\frac{\ln(x+1)}{x}=1$ and I think it expects me to use it somehow. How would you use it in this case? Is it similar to $\lim_{x\to0}\frac{\ln(x)}{x-1}=1$? – 2017-02-18
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    @MarkRead I made a small mistake with mine, as it should read $x\to1$. Note that if $x=u-1$, then$$\lim_{x\to0}\frac{\ln(x+1)}x=\lim_{u\to1}\frac{\ln(u)}{u-1}$$ – 2017-02-19
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$$\lim_{x\to1}\frac{\ln x}{x-1}=1$$ is a basic limit from high school, and expresses the logarithm is differentiable at $x=1$ and has derivative equal to $1$.