Question

lim as x-->0 of
xcscx
Can someone show steps please.

Answer 1

\[\lim_{x\rightarrow 0}[ x\csc(x)] =\lim_{x \rightarrow 0}[\frac{x}{\sin(x)}]\]
the second is indeterminate, 0/0. So, apply L'Hopitals:
\[\lim_{x\rightarrow 0} \frac{\frac{d}{dx}x}{\frac{d}{dx}\sin(x)}\implies \lim_{x \rightarrow 0} \frac{1}{\cos(x)}=\frac{1}{ \lim_{x \rightarrow 0}\cos(x)}=\frac{1}{1}=1\]

Answer 2

Thank you very much. I missed an important thing.

Answer 3

np

Answer 4

Can i ask you one more question

Answer 5

sure

Answer 6

wait, never mind i got it! Thanks

Answer 7

lol yw

Answer 8

if you know \[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1\]

then

\[\lim_{x\rightarrow 0} x\csc(x) =\lim_{x \rightarrow 0}\frac{x}{\sin(x)}=\lim_{x \rightarrow 0}\frac{1}{\frac{\sin(x)}{x}}=\frac{1}{1}=1\]

Answer 9

is this true for tanx/x=1?

Answer 10

\[\frac{\tan(x)}{x}=\frac{1}{\cos(x)}\frac{\sin(x)}{x}\]

Answer 11

\[\to \frac{1}{1}\cdot1=1\]

Answer 12

as \(x\to 0\)

Answer 13

hey @Zarkon can i ask you one more thing?

Answer 14

?

Answer 15

Can you explain this question: