MCQ
The value of $\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{{e^x} - 1}}{x}} \right)$ is
  • A
    $1/2$
  • B
    $\infty $
  • $1$
  • D
    $0$

Answer

Correct option: C.
$1$
c
(c) $\mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{\left( {1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + ... - 1} \right)}}{x} = 1$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

The solution set of $x<5$ and $x \geq 2$ is...
Choose the correct answer. If M and N are any two events, the probability that at least one of them occurs is:
If the coefficient of $x ^{10}$ in the binomial expansion of $\left(\frac{\sqrt{x}}{5^{\frac{1}{4}}}+\frac{\sqrt{5}}{x^{\frac{1}{3}}}\right)^{60}$ is $5^{ k } l$, where $l, k \in N$ and $l$ is coprime to $5$ , then $k$ is equal to
If $\lim _{x \rightarrow 0} \frac{\alpha x e^{x}-\beta \log _{e}(1+x)+\gamma x^{2} e^{-x}}{x \sin ^{2} x}=10, \alpha, \beta, \gamma \in R$, then the value of $\alpha+\beta+\gamma$ is $......$
The least remainder when ${17^{30}}$ is divided by $ 5$ is
The locus of the centre of circle which cuts the circles ${x^2} + {y^2} + 4x - 6y + 9 = 0$ and ${x^2} + {y^2} - 4x + 6y + 4 = 0$ orthogonally is
The condition that the parabolas $y^2 = 4ax$ and $y^2 = 4c(x - b)$ have a common normal other than $x-$ axis ($a, b, c$ being distinct positive real numbers) is
The number of $5$ digit numbers that can be formed with digits $1,2,3,4,5,6$ if the number must include $1$ and $2$ is-
Starting at time $t=0$ from the origin with speed $1 ms ^{-1}$, a particle follows a two-dimensional trajectory in the $x$-y plane so that its coordinates are related by the equation $y=\frac{x^2}{2}$. The $x$ and $y$ components of its acceleration are denoted by $a_x$ and $a_y$, respectively. Then

$(A)$ $a_x=1 ms ^{-2}$ implies that when the particle is at the origin, $a_y=1 ms ^{-2}$

$(B)$ $a_x=0$ implies $a_y=1 ms ^{-2}$ at all times

$(C)$ at $t=0$, the particle's velocity points in the $x$-direction

$(D)$ $a_x=0$ implies that at $t=1 s$, the angle between the particle's velocity and the $x$ axis is $45^{\circ}$

If the circles $x^2+y^2=a$ and $x^2+y^2-6 x-8 y+9=0$, touch externally, then $a=$