If f(x) = e^ 2x - (1 + 4x) ^ 1/2 (1 - x^2 ) for x 0, then f has - Vidyadip

MCQ

If $f(x) = \frac{{{e^{2x}} - {{(1 + 4x)}^{1/2}}}}{{\ln (1 - {x^2})}}$ for $x \ne 0,$ then $f$ has

A
an irremovable discontinuity at $x = 0$
✓
a removable discontinuity at $x = 0$ and $f (0) = -4$
C
a removable discontinuity at $x = 0$ and $f (0) = -1/4$
D
a removable discontinuity at $x = 0$ and $f (0) = 4$

✓

Answer

Correct option: B.

a removable discontinuity at $x = 0$ and $f (0) = -4$

b
$f(x)=\frac{e^{2 n}-(1+4 x)^{1 / 2}}{\ln \left(1-x^{2}\right)}, x \neq 0$

$f\left(0^{+}\right)=f\left(0^{-}\right)$

$f\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{e^{2 h}-(1+4 h)^{h}}{\ln \left(1-h^{2}\right)}$

$f\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{\left(1+2 h+\frac{4 h^{2}}{2}+\frac{3 h^{3}}{6}+\cdots\right)-\left(1+2 h-2 h^{2}+\cdots\right)}{\left(-h^{2}-\frac{h^{4}}{2}-\frac{h^{6}}{3}-\cdots \cdots\right)}$

$=\quad-4$

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