\lim_{x\to0} {e^{x}-e^{-x}}/{\sinx}=

AP ICET 2017 Shift 2 Paper

Question : 135 of 200

Marks: +1, -0

\lim_{x→0}\;\;{e^{x}-e^{-x}}/{\sinx}=

Solution:

\lim_{x→}\;\;{e^{x}-e^{-x}}/{\sinx}

=2[\lim↙{x→0}{({e^{x}-e^{-x}}/{2})}/{\sinx}]=2\;\;{\sinx}/{\sinx}=2

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