Question

LIM (xcosx-sinx)/
x-->0 (xsin^2(x))

Answer 1

\[\lim_{x \rightarrow 0}\frac{ xcosx-sinx }{ xsin ^{2}x }\]

Answer 2

i know i hv to use l'hopitals, but its kicking me out

Answer 3

Okay see:

\[\large{\lim_{x \rightarrow 0} \cfrac{xcosx-sinx}{xsin^2{x}}}\]

\[\large{\lim_{x \rightarrow 0} \cfrac{cosx - xsinx - cosx}{2x\sin x\cos x + sin^2x}}\]

\[\large{\lim_{x \rightarrow 0} \cfrac{-x\sin x}{\sin{x}(2x\cos x + \sin x)}}\]

\[\large{\lim_{x \rightarrow 0} \cfrac{-x}{2x\cos x + \sin x}}\]

Now, you can divide by x in both numerator and denominator.

Note: I applied LH Rule and Product rule for differentiation (also known as Liebnitz Rule) in 2nd step.

Answer 4

thank you @vishweshshrimali5 . let me see if i will get it. thank you for your assistance

Answer 5

Np

Answer 6

Remember that after dividing you will have to use this formula:

\[\large{\lim_{x \rightarrow 0}\cfrac{sinx}{x} = 1}\]

Answer 7

so for sinx/x it will =1, plus 2cos0= 2, and you are ryt because the answer is -1/3!

Answer 8

Great work @MERTICH

Answer 9

i wouldn't have cracked this one. am just started with this concept, and I am enjoying every bit of it. Thank you once again!

Answer 10

Do not mention it. Glad I could help :)