STD 11 - 12. limits — JEE कक्षा 11 साइन्स — Question

==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{\left[ {1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + ....} \right] - \left[ {1 - \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} - ....} \right]}}{{\sin x}}$

==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{2\,\left[ {\frac{x}{{1!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + .............} \right]}}{{\sin x}}$

==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{2\,\left[ {1 + \frac{{{x^2}}}{{3!}} + \frac{{{x^4}}}{{4!}} + ...........} \right]}}{{\frac{{\sin x}}{x}}}$

==> $y = \frac{{\mathop {\lim }\limits_{x \to 0} 2\,\left[ {1 + \frac{{{x^2}}}{{2!}} + .......} \right]}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}}}$

==> $y = \frac{2}{1} = 2$

ट्रिक : $ L- $ हॉस्पीटल नियम का प्रयोग करे

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$

$ = \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$