\limit_{x\to0} \frac{tanx-x}{x^2 tanx} is equal to:

NIMCET 2013 Question Paper with solutions

Question : 42 of 120

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\limit↙{x→0}

{tanx-x}/{x^2 tanx}

is equal to:

Solution:

\lim↙{x→0}\;\;{\tan x-x}/{x^{2}\tan x}

=\lim↙{x→0}[{\sec^{2}x-1}/{x^{2}\sec^{2}x+2x\tan x}]

Applying L’ Hospital’s rule,

=\lim↙{x→0}[\;\;{1-\cos^{2}x}/{x^{2}+2x\sin x\cos x}]

=\lim↙{x→0}[\;\;{\sin^{2}x}/{x^{2}+2x\sin x\cos x}]

=\lim↙{x→0}[\;\;{\sin^{2}x{/}x^{2}}/{1+\;\;{2\sin x}/{2}⋅\cos x}]=\;\;{1}/{3}

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