Question

The value of ${lim}_{x\to \infty }\frac{x^5}{5^x}+x(e^{1/x}-e^{-1/x})=2$

• A. True
• B. False

Answer 1

The correct option is A True

${lim}_{x\to \infty }\frac{x^5}{5^x}+\frac{e^{\frac{1}{x}}-e^{\frac{1}{x}}}{\frac{1}{x}}$
$={lim}_{x\to \infty }\frac{x^5}{5^x}+{lim}_{y\to 0}\frac{e^y-e^{-y}}{y}$ (put y = 1/x)
= 0 + 2 = 2