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Let f be a differentiable function a $x = 1$ such that $f(1) = 1$, $f'(1) = 4$. Compute the following limits, or prove that they don't exist:

a. $\lim_{x\to 1} \frac{1-f(x)}{x-1}$

b. $\lim_{x\to 1} \frac{f^2(x)-f(x)}{x-1}$

I am quite divided on this. According to the definition of the derivative, I get that a = -4 and b = 4.

However, if I check the limit of $\lim_{x\to 1+}$ and $\lim_{x\to 1-}$ on b, I get that the limit are different (One is positive and the other is negative).

Which step am I doing wrong? What am I missing?

Thanks.

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    Just to be clear, what does $f^2(x)$ mean? – 2017-01-18
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    Its means f(x) * f(x) – 2017-01-18
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    can you use L'Hospital? – 2017-01-18
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    Sorry I didn't mention it in my original post. Currently, we can't use L'Hospital. We'll get to it next week :) – 2017-01-18
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    :-) Then you'll have lots of (tad bit careless) fun using L'Hospital's I think. – 2017-01-18
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    @Guangliang I don't think that's the problem. Probably more about how to intuitively understand the left and right limits, since the OP claims to already know the solution to be $4$. – 2017-01-18

2 Answers 2

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If $f(x)$ is differentiable near $x=1$, then it is continuous and

$$\lim_{x\to1}\frac{f^2(x)-f(x)}{x-1}=f(1)\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=f(1)f'(1)$$

Taking $x\to1^+$, one has

$$f^2(x)-f(x)>0,\quad x-1>0$$

since for $x>1$, $f'(1)>0\implies f(x)>f(1)\implies f^2(x)>f(x)\implies f^2(x)-f(x)>0$.

Likewise, the same argument for $x<1$ shows $f^2(x)-f(x)<0$ and we also have $x-1<0$, so the limit from the right left side is also positive.

Thus, the limit from both the left and right sides are positive.

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    Ahh I see. I had a mistake saying the limits are of different signs. So my initial answer of a = -4, b =4 is correct. Thanks! – 2017-01-18
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    @Pashtidot :-P no problem. Chain of similar problems huh? – 2017-01-18
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$$(a)\;\;\lim_{x\to 1} \frac{1-f(x)}{x-1}=-\lim_{x\to 1} \frac{f(x)-1}{x-1}=-\lim_{x\to 1} \frac{f(x)-f(1)}{x-1}=-f'(1)=-4$$

$$(b)\;\;\lim_{x\to 1} \frac{f^2(x)-f(x)}{x-1}=\lim_{x\to 1} \frac{f(x)(f(x)-1)}{x-1}$$ $$=\lim_{x\to 1} f(x)\cdot \lim_{x\to 1} \frac{f(x)-1}{x-1}\underbrace{=}_{f\text{ coninuous at }x=1}f(1)f'(1)=4$$

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    I don't think that's the problem. Probably more about how to intuitively understand the left and right limits, since the OP claims to already know the solution to be $4$. – 2017-01-18
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    Well, now if the OP follows my approach can also check that in this case, the distinction between $x\to 1^+$ and $x\to 1^-$ is irrelevant. – 2017-01-18