Example:
Find the value of k, for which
$ f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, \text{if} \ -1\le x<0 \\ \frac{2x+1}{x-1}, \text{if} \ 0\le x<1 \end{cases} $
is continuous at x=0.
Solution:
Given that,
\( f(x)=\begin{cases} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}, \text{if} \ -1\le x<0 \\ \frac{2x+1}{x-1}, \text{if} \ 0\le x<1 \end{cases} \)
Since, \( f(x) \) is continuous at \( x=0 \),
\( \displaystyle\lim_{x \to 0^-} f(x)=f(0)=\displaystyle\lim_{x \to 0^+} f(x) \ …\ (1) \)
Here, \( f(0)=\frac{2(0)+1}{0-1} \)
\( =\frac{1}{-1} \)
\( f(0)=-1 \ ………. \ (2) \)
Now, \( \displaystyle\lim_{x \to 0^-} f(x)=\displaystyle\lim_{x \to 0}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x} \)
\( =\displaystyle\lim_{x \to 0}\frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\times \frac{\sqrt{1+kx}+\sqrt{1-kx}}{\sqrt{1+kx}+\sqrt{1-kx}} \)
\( =\displaystyle\lim_{x \to 0}\frac{\{1+kx\}-\{1-kx\}}{x(\sqrt{1+kx}+\sqrt{1-kx})} \)\( =\displaystyle\lim_{x \to 0}\frac{2kx}{x(\sqrt{1+kx}+\sqrt{1-kx})} \)
\( =\displaystyle\lim_{x \to 0}\frac{2k}{\sqrt{1+kx}+\sqrt{1-kx}} \)
\( =\frac{2k}{\sqrt{1+0}+\sqrt{1-0}} \)
\( =\frac{2k}{2} \)
\( \therefore \displaystyle\lim_{x \to 0^-} f(x)=k \ ……. \ (3) \)
From (1), (2) and (3), we have
\( k=-1 \)
