Let $f$ be the function of domain $\mathbb{R}$, defined by:

$$f(x)= \left\{ \begin{array}{lcc} \frac{1-e^{3x}}{x} & x <0 \\ \\ \log_2(k+2), & x \ge 0, k \in \mathbb{R^+} \\ \\ \end{array} \right.$$

a. Show that $f(x)$ is continuous at $]-\infty,0[$

I know that for a function to be continuous at a point $a$ the following must be true:

  1. $f(a)$exists
  2. $\lim_{x \rightarrow a}f(x)$ exists
  3. $f(a) = \lim_{x \rightarrow a}f(x)$

However, how do I solve for a range rather than a point?

Firstly, how do I prove a function is continuous at a range analitically? Is it enough to just check for asymptotes?

Assuming that checking for asymptotes is enough I can see that the only asymptote is $y=0$, as $\lim_{x \rightarrow -\infty}f(x) = 0$ and so it must be continuous

Also, looking at the graph I can see that $f(x)$ is continuous at $x<0$.

Now, how do I solve limits with ranges and figure out 2. and 3.?

In summary:

  1. How do I solve limits where $x$ tends to a range rather than to a constant?
  2. How do I prove that a function exists in a range? Is that the same as checking if the function is continuous in that range?
  3. How do I check if a function is continuous at a range? Do I just check for asymptotes?

Also, quick optional question (you don't have to answer this one):

In my book question b. of this exercise reads:

b. Determine $k$ so that $f(x)$ is continuous.

I have determined k to be $-\frac{15}{8}$ which my book says is correct, but the problem specifies that $k \in \mathbb{R^+}$. If $k$ is a negative number, shouldn't the exercise read $k \in \mathbb{R}$ or $k \in \mathbb{R^-}$instead?

  • 1
    $\begingroup$ Your answer for part b looks correct. Must be a typo in the book or something, as $\frac{-15}{8}$ is the only value $k$ can take to keep the function continuous. $\endgroup$ – WaveX Mar 22 '17 at 19:10

Note that $f$ restricted on $]-\infty, 0[$ is a composite function. The function $g: x \mapsto 1 - e^{3x}$ is continuous (I suppose you can use the result that $x \mapsto e^{x}$ is continuous. If not, note that $e^{x+h} - e^{x} = e^{x}(e^{h}-1)$ and you can prove it with a simple epsilon-delta argument.). The function $h: x \mapsto x$ is continuous and never zero on $]-\infty, 0[$. So $f$ restricted on $]-\infty,0[$, the quotient of $g$ over $h$, is continuous too by the elementary theorem that the sum, difference, product, and (meaningful) quotient of two continuous functions is still continuous.

  • $\begingroup$ I just realized I typed in the wrong function. I'm very sorry. I'll edit $\endgroup$ – Mark Read Mar 22 '17 at 19:04
  • $\begingroup$ No problem :). It just so happened that the argument above still works. $\endgroup$ – Megadeth Mar 22 '17 at 19:07

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