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Find a non-zero value for the constant $k$ such that $f$ defined as below is continuous at $x = 0$.

$$f(x) = \begin{cases} \frac{\tan(kx)}{x}, \hspace{0.3cm}x< 0 \\ 3x + 2k^2, \hspace{0.3cm}x\geq0 \end{cases}$$

My attempt:

$$\lim_{x\rightarrow0^{-}}\frac{\tan(kx)}{ x} = \lim_{x \rightarrow 0^{-}} \frac{\sin(kx)}{ x\cos(kx)} = \lim_{x \rightarrow 0^{-}} \frac{k\sin(kx)}{(kx)}\frac{1}{\cos(kx)} = k\lim_{x \rightarrow 0^{-}} \frac{1}{\cos(kx)} = k$$

$$\lim_{x \rightarrow 0^{+}}3x+2k^2 = 2k^2$$

For continuity we must have that the limit on the right must be equal to the limit on the left, i.e, $k = 2k^2$, so

$$k(2k-1) = 0$$

Therefore, a non-zero value for constant $k$ such that $f$ is continuous on $x=0$ is $k = 1/2$.

I am not fully confident in my solution. Could someone tell me if i went down the right path and why? Or maybe tell me I am entirely wrong and lend a hand?

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    Please use MathJax formatting in the future, to improve readability. :) – 2017-02-16
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    Mathjax? is that a seperate program? If so I will gladly download it! – 2017-02-16
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    No, it's just a way of typing your questions so they look nice and math-y. If you hit "edit" under the answer currently posted, you can see how he or she formatted everything. Just don't actually change it! – 2017-02-16
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    BTW, I didn't mean to seem harsh. Your question is nice and welcome, it's just a little tough to wade through visually. Welcome to the site. – 2017-02-16
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    Thank you and no I appreciate all advice. – 2017-02-16
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    Thanks for the help everyone I think I am learning a little more about the formatting! – 2017-02-16
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    I'll say! Great job! – 2017-02-16

1 Answers 1

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It seems you are correct. To put it more concisely, Note that $$\lim_{x \to 0-} \frac{ \tan kx}{x}=\lim_{x \to 0-} \frac{ \tan kx}{kx} \times k=k$$And $$\lim_{x \to 0+} 3x+2k^2=2k^2$$ So $2k^2=k$, and snce $k \neq 0$ so $k=\frac{1}{2}$.

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    Thank you for the feedback! Sorry if my question is ill formatted i am still figuring out how to do the arrows, subscripts and fractions. :) – 2017-02-16
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    @NickPavini No problem. See [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) if you want to learn more about Mathjax. Add the dollar signs, then add `\tan x`. Then it will render as $\tan x$. – 2017-02-16
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    $\lim\limits_{x\to0^+} f(x) =3x+2k^2$ – 2017-02-16
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    @S.C.B. when i type $$\tan x will it automatically convert or will i visually see it as you posted? – 2017-02-16
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    @NickPavini automatically. You have to add it like this `$\tan x $`. – 2017-02-16
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    @NickPavini Could you accept my answer? Thank you. I don't mean to force you, but I would like to take this questin off the unanswered section. – 2017-02-16
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    @NickPavini And of course, upvote if it helped you, too! – 2017-02-16
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    @S.C.B. ya no problem. – 2017-02-16
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    @TheCount I cant upvote until I get 15 rep, or I would for sure! :) – 2017-02-16
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    Ahh, @NickPavini That's true, I forgot about that. Well, in due time. – 2017-02-16
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    @NickPavini Good luck, but this comment section is unsuited for that sort of discussion. Also, You could gain the association bonus if you gain 85 more reputation on stackoverflow, I believe. – 2017-02-16