lim-x-0-5e-x-5e-x-10-4x-2-

Question Number 168014 by mathlove last updated on 31/Mar/22

\lim_{x \to 0} \frac{5 e^{x} + 5 e^{- x} - 10}{4 x^{2}} = ?

Answered by som(math1967) last updated on 01/Apr/22

\frac{5 e^{x} + \frac{5}{e^{x}} - 10}{4 x^{2}}

= \frac{5 e^{x} - 5 - \frac{5}{e^{x}} (e^{x} - 1)}{4 x^{2}}

= \frac{5 (e^{x} - 1) - \frac{5}{e^{x}} (e^{x} - 1)}{4 x^{2}}

= \frac{(e^{x} - 1) (5 - \frac{5}{e^{x}})}{4 x^{2}}

= \frac{5 (e^{x} - 1) (e^{x} - 1)}{4 x^{2} e^{x}}

∴ \underset{x \to 0}{l i m} \frac{5}{4} \times {(\frac{e^{x} - 1}{x})}^{2} \times \frac{1}{e^{x}}

= \frac{5}{4} \times 1^{2} \times 1 = \frac{5}{4}

Commented by som(math1967) last updated on 01/Apr/22

o t h e r w a y

\frac{5}{4} \underset{x \to 0}{l i m} \frac{e^{x} + e^{- x} - 2}{x^{2}}

= \frac{5}{4} \underset{x \to 0}{l i m} \frac{e^{2 x} + 1 - 2 e^{x}}{e^{x} \times x^{2}}

= \frac{5}{4} \underset{x \to 0}{l i m} {(\frac{e^{x} - 1}{x})}^{2} \times \frac{1}{e^{x}}

= \frac{5}{4} \times 1^{2} \times 1 = \frac{5}{4}