-4
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For what value of k,

$$f(x)=\begin{cases}\frac{\sqrt{1+kx}-\sqrt{1- kx}}{x} & \mbox{ if }-1 \le x <0 \\ \frac{2x+1}{x-1} & \mbox{ if } 0\le x<1\end{cases}$$

is continuous at $x= 0$.

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    $\begingroup$ What have you tried? Have you tried to calculate the limit $\lim_{x\to0^-}f(x)$? $\endgroup$ – skyking Jan 14 '16 at 15:11
  • $\begingroup$ Yes, I tried it and my answer came -1 . Is it correct? $\endgroup$ – Anubhav Goel Jan 14 '16 at 15:15
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    $\begingroup$ No, thats the right hand limit. For the left hand limit you would have to use the upper expression. $\endgroup$ – skyking Jan 14 '16 at 15:17
  • $\begingroup$ My answer to 0/0 form came k. Is it right? $\endgroup$ – Anubhav Goel Jan 14 '16 at 15:32
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    $\begingroup$ If you mean you found $k$ to be $-1$, then: yes, correct! $\endgroup$ – StackTD Jan 14 '16 at 15:33
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For $f$ to be continuous at $x=0$, $f$ needs to be defined at $x=0$ and you need $$\lim_{x \to 0} f(x) = f(0)$$ Clearly $f(0) = -1$ and the right hand limit (using the same part of the function) gives you -1 as well. You need to calculate the left hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1+kx}-\sqrt{1- kx}}{x} = \ldots$$ Can you calculate this limit? It will depend on the parameter $k$ and you need this limit to be equal to $-1$, solve for $k$.

To evaluate the limit, you could use the following trick: $$\frac{\sqrt{1+kx}-\sqrt{1- kx}}{x} = \frac{\left(\sqrt{1+kx}-\sqrt{1- kx}\right)\left(\sqrt{1+kx}+\sqrt{1- kx}\right)}{x\left(\sqrt{1+kx}+\sqrt{1- kx}\right)} = \ldots$$ In the numerator, use $(a-b)(a+b)=a^2+b^2$ and simplify. Can you take it from here?

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  • $\begingroup$ I used very same trick. $\endgroup$ – Anubhav Goel Jan 14 '16 at 15:27
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    $\begingroup$ Alright; what value for $k$ did you get? $\endgroup$ – StackTD Jan 14 '16 at 15:31

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