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Differential Calculus> Can u solve it? Question number 13 plz...

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Can u solve it? Question number 13 plz

fhhgv , 10 Years ago

Grade 11

1 Answers

Riddhish Bhalodia

By expanding the tan2x and cos2x terms we get

$L = lim_{x\rightarrow 0} \frac{x(\frac{2tan(x)}{1-tan^2(x)} -2tan(x))}{4sin^4(x)}$
simplifying we get

$L = 0.5 lim_{x\rightarrow 0}\frac{x}{sin(x)}\frac{1}{cos^3(x)}\frac{1}{1-tan^2(x)}$
now each of the multiplying term has limit 1 as x tends to 0 thus
L = 1/2

Last Activity: 10 Years ago

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