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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 $k$ = $-\frac{11}{3}$

Robangiu
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2 Answers2

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$k = -11/3 $ is correct. But I do not know what you have done, since you conceal your work !

Fred
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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.

user209663
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