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Let $$ h(x) = \begin{cases} \sin (kx), & \text{if }x\leq 2, \\ x+k^2, & \text{if }x>2, \end{cases} $$ where $k$ is a real constant. Determine the value of $k$, if any, that makes k continuous everywhere.

I done my work and here goes.

Since $h(x)$ is continuous on $(-\infty,2)$ and on $(2,\infty)$, it suffice to find the value of $k$ such that $h(x)$ is continuous at $x=2$.

Suppose $h(x)$ is continuous at $x=0$, then $$\lim_{x\to2^-}h(x)=\lim_{x\to2^-}\sin k(x)=\sin(2k)$$ and $$\lim_{x\to2^+}h(x)=\lim_{x\to2^+}x+k^2=2+k^2$$ Therefore, $\lim_{x\to2^+}h(x)=\lim_{x\to2^-}h(x)$ implies $\sin(2k)=2+k^2$

Ok so now I am stuck. Thanks in advance for helping me out!

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$|\sin x|\leq 1$ for any real $x$, so there are no such $k$ –  pritam Sep 25 '12 at 16:40

2 Answers 2

up vote 1 down vote accepted

Your work so far is fine.

Hint: $\sin(-)$ is always between -1 and 1.

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Since $1\geq\sin(2k)\geq-1$, then $1\geq2+k^2\geq-1=-1\geq k^2\geq-3$. Therefore, there is no such k. –  Yellow Skies Sep 25 '12 at 16:47
    
Thanks! That helped. –  Yellow Skies Sep 25 '12 at 16:50

$-1\leq \sin(2k)\leq 1$ and $2+k^2\geq 2$

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