Let [t] denote the greatest integer $\le$ t
and $\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right] = A$.
Then the function,
f(x) = [x2]sin($\pi$x) is discontinuous, when x is
equal to :
Solution
A = $\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)$$
<br><br>= 4
<br><br>Now, when x = $\sqrt {A + 1}$ = $\sqrt 5$, f(x) = [x<sup>2</sup>]sin($\pi$x) is discontinuous at this non integer point.
<br><br>But at x = 2, 3 and 5, f(x) is continuous.
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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