Let f(x) = ,and n be number of points of discontinuity of the function . The value of is equal to.

JEE MainMathematicsModerateLimits, continuity and derivability

Let f(x) = $\frac{1}{x - 1}$ ,and n be number of points of
discontinuity of the function $f\left( \frac{1}{x - 1} \right)$ . The
value of
$\lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{n}} - \frac{cotx}{x} \right)$
is equal to.

Show answer and solution

f(x) = \frac{1}{x - 1}

is discontinuous at x = 1

f\left( \frac{1}{x - 1} \right) = \frac{1}{\frac{1}{x - 1} - 1}

\frac{1}{x - 1} = 1 \Rightarrow 1 = x - 1 \Rightarrow x = 2

\therefore

x = 1, 2 all two points of discontinuities

\Rightarrow n = 2

\lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{n}} - \frac{\cot x}{x} \right) = \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{2}} - \frac{\cot x}{x} \right)

= \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{2}} - \frac{1}{x}\left( \frac{1}{x} - \frac{x}{3} + \ldots\ldots \right) \right)

= \lim_{x \rightarrow 0}\mspace{2mu}\left( \frac{1}{x^{2}} - \frac{1}{x^{2}} + \frac{1}{3} + \ldots\ldots \right) = \frac{1}{3}

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

Scan any doubt from your notes and get a whiteboard explanation.