f(n) = \begin{cases} \frac{ln(3x+1)}{5x}, & \text{ $x$ >0} \\[2ex] \frac{2x^2 -1}{k+2}, & \text{if $x$ $\leq$0} \end{cases}
The value I found is $k$ =$-\frac{11}{3}$ but I'm not sure I did it right, here's what I did:
$\lim\limits_{x \to 0} \frac{ln(3x^2-1)}{5x}$ = $\lim\limits_{x \to 0} \frac{(2x^2-1)}{k+2}$
and since $\lim\limits_{x \to 0} \frac{ln(3x^2-1)}{5x}$ equals $\frac{3}{5}$,
$\frac{3}{5}$=$\frac{2x^2-1}{k+2}$ then I multiplied by 5 and then by $k$+2 to remove the fraction,
3$k$+6 = $10x^2$-5, as stated by the limit the x is = 0, so resolving a little bit I find:
3$k$ = -11 and of course $k$ = $-\frac{11}{3}$
Yes, that is correct. Here is why:
For $f(x)$ to be continuous at $x=0$, we must have: $\lim_{x \rightarrow 0} f(x) = f(0)$.
From the way you defined $f(x)$, we have: $f(0) = \dfrac{-1}{k+2} $
Now, the limit as $x \rightarrow 0^-$ (from the left) is just $$\lim_{x \rightarrow 0^-} f(x) = \dfrac{-1}{k+2} $$
so now we need to consider the right-side limit $(x \rightarrow 0^+)$: $$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \dfrac{\ln(3x+1)}{5x} = \lim_{x \rightarrow 0^+} \dfrac{ \dfrac{3}{3x+1} }{5} = \dfrac{3}{5} $$ Note that I used the L'Hospital rule to get that limit.
Now to get the $k$ that satisfy the continuity condition, we set $$ \dfrac{-1}{k+2} = \dfrac{3}{5}$$ Then solve for $k$, which will give you $k = -\dfrac{11}{3}$
Hope that help.