MCQ
$\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$ is
- A$0$
- B$1$
- ✓$2$
- DNon existent
==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{\left[ {1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + ....} \right] - \left[ {1 - \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} - ....} \right]}}{{\sin x}}$
==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{2\,\left[ {\frac{x}{{1!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + .............} \right]}}{{\sin x}}$
==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{2\,\left[ {1 + \frac{{{x^2}}}{{3!}} + \frac{{{x^4}}}{{4!}} + ...........} \right]}}{{\frac{{\sin x}}{x}}}$
==> $y = \frac{{\mathop {\lim }\limits_{x \to 0} 2\,\left[ {1 + \frac{{{x^2}}}{{2!}} + .......} \right]}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}}}$
==> $y = \frac{2}{1} = 2$
Trick : Applying $L-$ Hospital’s rule,
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + {e^{ - x}}}}{{\cos x}} = \frac{{{e^0} + \frac{1}{{{e^0}}}}}{{\cos 0}} = \frac{{1 + 1}}{1} = 2$.
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$x^2+y^2+5 x-3 y+4=0 .$
Then which of the following statements is (are) TRUE?
$(A)$ $\alpha=-1$ $(B)$ $\alpha \beta=4$ $(C)$ $\alpha \beta=-4$ $(D)$ $\beta=4$
- the vowels occur in the same order $(EUAIO)$,
- the consonants occur in the same order $(DCTN)$,
- no two consonants are next to each other.