If f(x) = begin{cases} frac{sqrt{1+mx} - sqrt | vedclass

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(B) Since $f(x)$ is continuous in $[-1, 1]$,it must be continuous at $x = 0$. Therefore,$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}}

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(B) Since $f(x)$ is continuous in $[-1, 1]$,it must be continuous at $x = 0$. Therefore,$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{+}} f(x) = f(0)$. First,calculate the right-hand limit: $\lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} \frac{2x+1}{x-2} = \frac{2(0)+1}{0-2} = -\frac{1}{2}$. Next,calculate the left-hand limit: $\lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} \frac{\sqrt{1+mx} - \sqrt{1-mx}}{x}$. Rationalizing the numerator: $\lim_{x \rightarrow 0^{-}} \frac{(\sqrt{1+mx} - \sqrt{1-mx})(\sqrt{1+mx} + \sqrt{1-mx})}{x(\sqrt{1+mx} + \sqrt{1-mx})} = \lim_{x \rightarrow 0^{-}} \frac{(1+mx) - (1-mx)}{x(\sqrt{1+mx} + \sqrt{1-mx})} = \lim_{x \rightarrow 0^{-}} \frac{2mx}{x(\sqrt{1+mx} + \sqrt{1-mx})} = \frac{2m}{1+1} = m$. Equating the limits: $m = -\frac{1}{2}$.

If $f(x) = \begin{cases} \frac{\sqrt{1+mx} - \sqrt{1-mx}}{x}, & -1 \le x < 0 \\ \frac{2x+1}{x-2}, & 0 \le x \le 1 \end{cases}$ is continuous in the interval $[-1, 1]$,then $m$ is equal to:

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