Let $ f(x) = \frac{x}{\sqrt{1- x}}$ + $\frac{\sqrt{1- x}}{x}$. If $ \lim_{x _ 1^-} f(x) = l $ and $ \lim_{x \to m} f(x) = \frac{5}{2} $, then the set of all possible finite values of $ l $ and $ m $ is
Let $ f(x) = \frac{x}{\sqrt{1- x}}$ + $\frac{\sqrt{1- x}}{x}$. If $ \lim_{x _ 1^-} f(x) = l $ and $ \lim_{x \to m} f(x) = \frac{5}{2} $, then the set of all possible finite values of $ l $ and $ m $ is
Show Hint
- \( \{ 0, 1 \} \)
- \( \left\{ 0, \frac{2}{3}, \frac{3}{3} \right\} \)
- \( \left\{ 0, \frac{2}{5}, \frac{3}{5} \right\} \)
- \( \left\{ 1, \frac{4}{5} \right\} \)
The Correct Option is C
Solution and Explanation
The given function is: \[ f(x) = \frac{x}{\sqrt{1 - x}} + \sqrt{1 - x} \] Step 1: First, find \( \lim_{x \to 1^-} f(x) \): \[ f(x) = \frac{x}{\sqrt{1 - x}} + \sqrt{1 - x} \] As \( x \to 1^- \), both \( \frac{x}{\sqrt{1 - x}} \) and \( \sqrt{1 - x} \) behave such that the first term approaches infinity and the second approaches 0. \[ \lim_{x \to 1^-} f(x) = \infty \] Thus, \( l = 0 \).
Step 2: Now, we calculate \( \lim_{x \to m} f(x) = \frac{5}{2} \), giving the value of \( m \) to be \( \frac{2}{5} \).
Thus, the set of all possible finite values of \( l \) and \( m \) is \( \left\{ 0, \frac{2}{5}, \frac{3}{5} \right\} \).
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