Question

lim x=0 xsinx/2-2cosx

Answer 1

can you use LHopital's?

Answer 2

type the equation properly.

Answer 3

lim x=0 (xsinx)/(2+2cosx)

Answer 4

2+2cosx? are u kidding me?

Answer 5

u wrote 2-2cosx in the question

Answer 6

o i mean 2-2cosx sorry

Answer 7

\[\Huge \frac{x sinx}{2(1+cosx)}=\frac{2xsin \frac{x}{2}\cos{\frac{x}{2}}}{2(2\cos^2 \frac{x}{2})}\]

\[\Huge \frac{\cancel{2}x \{\sin \frac{x}{2}\cancel{cos{\frac{x}{2}}}}{\cancel{2}(2\cancel{\cos^2 \frac{x}{2}})}\]
\[\Huge \frac{1}{2} \tan \frac{x}{2}\]
=0 if x->0
Well if it is 2+2cosx

Answer 8

okay nevermind

Answer 9

yr its 2-2cosx

Answer 10

\[\LARGE \frac{xsinx}{2-2cosx}=\frac{x(x-\frac{x^3}{6})}{2-2(1-\frac{x^2}{2})}=\frac{x^2}{x^2}=1\]
Using series expansions

Answer 11

final answer is 1 if you use LHopital's rule.

Answer 12

neglect x^3/6 and other higher order terms

Answer 13

when x->0 they tend to 0

Answer 14

@dls series answer what you have given above is right answer is 1 but previously xtan(x/2)
answer is zero which is wrong.

Answer 15

read properly,he wrote 2+2cosx in the previous reply,so I changed my solution to 2+2cosx in denominator

Answer 16

what ever it is ,series solution is right.