Question
$\lim\limits_{\text{x} \rightarrow0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}$ is equal to:
-
$\text{n}$
-
$1 $
-
$-\text{n}$
-
$0$
$\lim\limits_{\text{x} \rightarrow0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}$ is equal to:
$\text{n}$
$1 $
$-\text{n}$
$0$
Solution:
Given $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-1}{\text{x}}=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{(1+\text{x})-(1)}$
$=\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{\text{n}}-(1)^{\text{n}}}{1+\text{x}-(2)}=\text{n}(1)^{\text{n}-1}$
$=\text{n}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
The distance between the foci of a hyperbola is 16 and its eccentricity is 2. Its equation is:
$\text{x}^2-\text{y}^2=32$
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
$2\text{x}-3\text{y}^2=7$
none of these.
| Column-I | Column-II | ||
| (A) | If $\text{P}(\text{n},4)=20.\text{P}(\text{n},2)$ then the value of n is | (1) | 28 |
| (B) | $\ ^5\text{p}_\text{r}=\ ^{26}\text{p}_\text{r-1}$ | (2) | 4 |
| (C) | $\ ^5\text{p}_\text{r}=\ ^{6}\text{p}_\text{r-1}$ | (3) | 7 |
| (D) | Value of $\frac{8!}{6!\times2!}$ is | (4) | 3 |