Question

The value of $\underset{x\to 0^{-}}{lim}\frac{\left[x\right]+\left[x^2\right]+\left[x^3\right]+....+\left[x^{2n+1}\right]+n+1}{1+\left[x^2\right]+|x|+2x}$, n $\in$ N is equal to?

• A. $n+1$
• B. $n$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

When $x\to 0^{-}$, we have
$\left[x^{2n+1}\right]=-1$ and $\left[x^{2n}\right]=0$ for $n=0,1,2,....$
$\therefore \underset{x\to 0^{-}}{lim}\frac{\left[x\right]+\left[x^2\right]+\left[x^3\right]+....+\left[x^{2n+1}\right]+n+1}{1+\left[x^2\right]+|x|+2x}$
$=\underset{x\to 0^{-}}{lim}\frac{(-1)+0+(-1)+0+....+0+(-1)+n+1}{1+0-x+2x}$
$=\underset{x\to 0^{-}}{lim}\frac{(-1)(n+1)+n+1}{1+x}$
$=\frac{0}{1+0}=0$