Calculus - limits

1357

(1-x²cot²x)/x² Applying L'Hospital
=-cot²x+xcotxcosec²x
=(cosx/sinx)(x/sin²x - cosx/sinx)
=(cosx/sin³x)(x-sinxcosx)=(cosx/sin³x)(x-sin2x/2)
Applying limits wherever possible and then applying L Hospital

(1-cos²x)/3sin²x=2sin²x/3sin²x=2/3

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

sindhu br ·Jul 13 '09 at 0:47

(1-x²cot²x)/x²
L'Hospital cant be applied becuz numerator and denominator are not tending to 0

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

yes no ·Jul 13 '09 at 1:07

@above
arey simplify it na in terms of cos and sin..u will see that its a 0/0 form

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

gordo ·Jul 13 '09 at 1:27

u can also use series expansion to do it,

[sin²x-x²cos²x]/x²sin²x

we have $[\left(x-\frac{x^{3}}{3!} \right)^{2}-x^{2}\left(1-\frac{x^{2}}{2!} \right)^{2}}]/[x^{2}\left(x-x^{3}/3! \right)^{2}]$

applying binomial theoram,

= $[x 2 (1 - 2 x 2 / 3!) - x 2 (1 - 2 x 2 / 2!)] / [x 4 (1 - 2 x 2 / 3!)]$

simplifying this we have 2/3 as the final answer

cheers!!

UP 0 DOWN 0
Post on FacebookPost anonymously?

1357

Manish Shankar ·Jul 13 '09 at 2:20

[(1-(x/tanx)²]/x²

I considered (x/tanx) tending to 1 and hence applied L'Hospital

UP 0 DOWN 0
Post on FacebookPost anonymously?

limits

5 Answers

Your Answer

Similar Questions

limits

5 Answers

Your Answer

Add Image

Similar Questions