\lim_{x\to0} {e^{3x}-e^{4x}}/{\sin3x\cos4x}=

AP ICET 2016 Paper

Question : 112 of 200

Marks: +1, -0

\lim↙{x→0}\;\;{e^{3x}-e^{4x}}/{\sin3x\cos4x}=

Solution:

Given expression :

\lim↙{x→0}\;{e^{3x}-e^{4x}}/{\sin3x.\cos4x}

=\lim↙{x→0}{({e^{3x}-1}/{x})-({e^{4x}-1}/{x})}/{3({\sin3x}/{3x})⋅\cos4x}

={3-4}/{3}={-1}/{3}

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