Define $f(x)=\begin{cases} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2…

Define $f(x)=\begin{cases} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & \text{if } -1 \leq x < 0 \\ \frac{2 x+1}{x-2}, & \text{if } 0 \leq x \leq 1 \end{cases}$ If $\lim_{x \rightarrow 0} f(x)$ exists, then $p=$
  1. -1
  2. \(-\frac{1}{2}\)
  3. \(\frac{1}{2}\)
  4. 1

Solution

Given, \(f(x)=\left\{\begin{array}{cl}\frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}, & \text { if }-1 \leq x < 0 \\ \frac{2 x+1}{x-2}, & \text { if } 0 \leq x \leq 1\end{array}\right.\) Now, RHL \(=\) LHL (at \(x=0\)) \(\Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{2 x+1}{x-2}=\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+p x}-\sqrt{1-p x}}{x}\) Apply L'Hospital rule in LHL \(\Rightarrow \quad-\frac{1}{2}=\lim _{x \rightarrow 0^{-}} \frac{\frac{p}{2 \sqrt{1+p x}}+\frac{p}{2 \sqrt{1-p x}}}{1}\) \(\Rightarrow \quad-\frac{1}{2}=\frac{p}{2}+\frac{p}{2} \Rightarrow-\frac{1}{2}=p \Rightarrow p=-\frac{1}{2}\)

Asked in: AP EAMCET 2019 (22 Apr Shift 1)

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