Question

Which of the following function is discontinuous at x = 0, if f(0) = 1

$f(x) = \frac{sin(sin^2x)}{(e^{|sinx|}-1)\sqrt{x^2}}$; (x$\neq$0)

$f(x) = (1+x)(sgn(x)) +x²$; (x$\neq$0) (where sgn(x) is signum function)

$f(x) = (In(1 + tan²x)) (cosec((e\*-1)²)); (x$\neq$ 0)

$f(x) = (1 + sin²x) cosecx; (x$\neq$0)

• A. f(x) = $\frac{sin(sin^2x)}{(e^{|sinx|}-1)\sqrt{x^2}}$; (x$\neq$0)
• B. f(x) = (1+x)(sgn(x)) +x²; (x$\neq$0) (where sgn(x) is signum function)
• C. f(x) = (In(1 + tan²x)) (cosec((e*-1)²)); (x$\neq$ 0)
• D. f(x) = (1 + sin²x) cosecx; (x$\neq$0)

Answer 1

Answer: (B), (C), (D)

Options 2, 3, and 4 are discontinuous at x = 0.

Answer 2

Option 1:
For small $x$,
$\sin(\sin^2x) \approx x^2$.
Also, $e^{|\sin x|}-1 \approx |x|$ and $\sqrt{x^2} = |x|$.
Hence,

$$f(x) \approx \frac{x^2}{|x|\cdot|x|} = \frac{x^2}{x^2} = 1.$$

So $\lim_{x\to0} f(x) = 1 = f(0)$ (Continuous).

Option 2:
$f(x)=(1+x)\operatorname{sgn}(x)+x^2$.

• For $x>0$: $\operatorname{sgn}(x)=1$, so $f(x)=1+x+x^2$ with $\lim_{x\to0^+} f(x)=1$.
• For $x<0$: $\operatorname{sgn}(x)=-1$, so $f(x)=-1-x+x^2$ with $\lim_{x\to0^-} f(x) = -1$.
  Since the one–sided limits are different, $f(x)$ is discontinuous at $0$.

Option 3:
$f(x)=\ln(1+\tan^2x)\cdot \csc((e^{-1})^2)$ (here, $\csc((e^{-1})^2)$ is a nonzero constant because $\sin((e^{-1})^2) \neq 0$).
For small $x$, $\tan x \approx x$ so $\ln(1+\tan^2x) \approx x^2$. Hence,

$$\lim_{x\to0} f(x) \approx \csc((e^{-1})^2)\cdot x^2 = 0.$$

Since $f(0)=1$ but the limit is 0, it is discontinuous at $0$.

Option 4:
$f(x)=(1+\sin^2x)\csc x = \frac{1+\sin^2x}{\sin x}$.
As $x\to0$, the denominator $\sin x \to 0$ while the numerator tends to 1, so $f(x)$ blows up (tends to $\infty$ or $-\infty$).
Hence, it is discontinuous at $0$.