MCQ
$\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}}} \right\} = $
- ✓$1/120$
- B$-1/120$
- C$1/20$
- DNone of these
Aliter : Apply $L-$ Hospital’s rule
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x - 1 + \frac{{3{x^2}}}{6}}}{{5{x^4}}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \sin x + \frac{{6x}}{6}}}{{20{x^3}}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \cos x + 1}}{{60{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{120\,x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{120}} = \frac{1}{{120}}.$
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$(A)$ $\arg (-1- i )=\frac{\pi}{4}$, where $i =\sqrt{-1}$
$(B)$ The function $f: R \rightarrow(-\pi, \pi]$, defined by $f(t)=\arg (-1+i t)$ for all $t \in R$, is continuous at all points of $R$, where $i=\sqrt{-1}$
$(C)$ For any two non-zero complex numbers $z_1$ and $z_2$, $\arg \left(\left(\frac{z_1}{z_2}\right)-\arg \left(z_1\right)+\arg \left(z_2\right)\right.$ is an integer multiple of $2 \pi$.
$(D)$ For any three given distinct complex numbers, $z_1, z_2$ and $z_3$, the locus of the point $z$ satisfying the condition $\arg \left(\frac{\left( z - z _1\right)\left( z _2- z _3\right)}{\left( z - z _3\right)\left( z _2- z _1\right)}\right)=\pi$, lies on a straight line