Show a function is well-defined and continuous

Question

We have a function $f(x) = \sqrt{x^2 + x + 4} + x$ for $x \leq -1$, and $f(x) = \dfrac{2x^3 - 2x^2 - 4x}{(x^2-1)(x-2)}$ for $x \geq -1 ( x \neq 1,2$). Show that $f$ is well defined and continuous on $(- \infty, -1)$. Also, determine whether or not $f$ is continuous in $-1$.

We have to show that there exists a $\delta$ for every $\epsilon$, such that if $|x-x_0| < \delta$, then $|f(x)-f(x_0)| < \epsilon$.

So $| (\sqrt{x^2 + x + 4} + x) - (\sqrt{x_0^2 + x_0 + 4} + x_0) | < \epsilon $.

Now we have to find an expression of $\epsilon$ in terms of $\delta$.

• The first part of this question is how to do this, because, algebraically, I can't figure it out.

The second part of the question is regarding.

Also, determine whether or not $f$ is continuous in $1$.

I took the liberty of looking at this question and trying to answer it:

$f(-1) = 1$. We also know that $ \lim_{x \to -1} \dfrac{2x^3 - 2x^2 - 4x}{(x^2-1)(x-2)} = 1$. Hence $f(x)$ is continuous in $-1$

• Is this correct?

Answer 1

It is well defined on $(-\infty, -1)$ as $x^2+x+4$ is always non negative (I'm sure you can see why). You can argue continuity on $(-\infty, -1)$ by considering for example the composition of $\sqrt {x}$ and $x^2+x+4$ which are both continuous.

At $-1$ you will have to consider the right and left limits, check if they are equal and if they are, whether their value is $f(-1)$ For the right limit notice that the denominator becomes $0$ for $x=-1$ but you can simplify the fraction by factoring out $x+1$. You can do that because to calculate the limit at $-1$ you don't need your function to be defined at $-1$

Answer 2

Hints.

• You need to show it is well-defined on $(-∞,-1)$ first. For continuity, it might be easier to argue based on some known simple continuous functions, rather than the $\epsilon-δ$ definition.
• You mistyped 'continuous at $\color{red}{-}1$'. your "we also know" is not sufficient if you want any marks: show why you know it.