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Determine the values of $k$ for which the function $f$, defined by

$$f(x)= \begin{cases} \dfrac{1}{x-1}, & x<1 \\ \\ \dfrac{1}{k(x-1)}, & x>1 \end{cases} $$

has a limit when $x \rightarrow 1$

I did:

  • I know that in order for this limit to exist, $\lim_{x\rightarrow1^+}f(x)$ and $\lim_{x\rightarrow1^-}f(x)$ have to exist and be different from infinity (I think)
  • Since $f(x)$ where $x < 1$ doesn't mention $k$, I assume that the $(x-1)$ in $\frac{1}{k(x-1)}$ is always a positive value
  • Therefore, the sign of $k$ will determine the sign of $\frac{1}{k(x-1)}$
  • I know that $k$ can't be equal to 0 because $\frac{1}{k(x-1)}$ would become impossible to solve

I tried plotting $\frac{1}{k(x-1)}$ with several values for $k$, and basically what I get is that the larger the absolute value, the smaller the function and changing the sign simply makes a vertical reflexion of the function.

My book says the solution is $k < 0$.

I don't understand why. Can someone explain this to me?

1 Answers 1

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I believe your book wants the limit to be 'the same' when $x\to1$? Because $f\to -\infty$ as $x\to1^{-}$, so there can be no finite limit.

When $k<0$, $f$ also goes to $-\infty$ as $x\to1^+$. When $k>0$, $f$ goes to $+\infty$ as $x\to1^+$, so the left and right limits do not agree.

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    Maybe this question doesn't make any sense. What a mess. In an earlier question, my book asks me to prove that a function doesn't have a certain limit by showing it tends to infinity as x tends to the given value. So yeah, there is a possibility that this question doesn't make sense. – 2017-02-11