Question

The value of f(0) so that f(x) = $\frac{- e^{x} + 2^{x}}{x}$is continuous at

x = 0, will be

• A. – ln 2
• B. 0
• C. 4
• D. ln$\left( \frac{2}{e} \right)$

Answer 1

Answer: (D)

ln$\left( \frac{2}{e} \right)$

Answer 2

$\lim _ { x \rightarrow 0 }$ $\left( \frac{- e^{x} + 2^{x}}{x} \right)$=$\lim _ { x \rightarrow 0 }$ $\left( \frac{- (e^{x} - 1)}{x} + \frac{(2^{x} - 1)}{x} \right)$

= ln2 – lne = ln(2/e)