Question

$lim \\ x - 0 |8 ^{sinx} - 2 {}^{tanx} \div e {}^{2x} - 1 |$
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Answer 1

Answer:

limx->0  8^sinx - 2^tanx / e^2x -1

lim x->0 {8^sinx -1/x - (2^tan x - 1)/x}/e^2x-1/x

= ln 8 - ln 2 / 2

= ln 4 / 2

= 2ln 2 / 2 = ln 2

Hope this helps you !