Question

Which of the following is an indeterminate form (where $[.]$ denotes greatest integer function)

• A. $\underset{x\to \infty }{lim}\frac{\sin x}{x}$
• B. $\underset{x\to 0^{+}}{lim}\frac{[x{]}^2}{x^2}$
• C. $\underset{x\to 1}{lim}\frac{x-1}{x^3-1}$
• D. $\underset{x\to \infty }{lim}\sqrt{x^4-1}+x$

Answer 1

The correct option is C $\underset{x\to 1}{lim}\frac{x-1}{x^3-1}$

$\left(A\right)-1\le \sin x\le 1$
$\therefore \underset{x\to \infty }{lim}\frac{\sin x}{x}=\frac{\text{finite}}{\infty }=0$
Hence, it's not an indeterminate form.

$\left(B\right)x\to 0^{+},\left[x\right]=0$
$\underset{x\to 0^{+}}{lim}\frac{[x{]}^2}{x^2}=\underset{h\to 0}{lim}\frac{[0+h{]}^2}{h^2}=\frac{0}{h^2}=0$
Hence, it's not an indeterminate form.

$\left(C\right)\underset{x\to 1}{lim}\frac{x-1}{x^3-1}=\frac{0}{0}$
Hence, it's an indeterminate form.

$\left(D\right)\underset{x\to \infty }{lim}\sqrt{x^4-1}+x=\sqrt{(\infty {)}^4+1}+\infty$
$=\infty +\infty =\infty$
Hence, it's not an indeterminate form.