Question

Find the limit of the trigonometric function.
Limit as x approaches 0 is sinx(1-cosx)/x

Answer 1

\[\lim_{x \rightarrow 0} \frac{ sinx }{ x }*(1-cosx)= 1*(1-1)=0\]

Answer 2

why does (1-cosx) become 1

Answer 3

it doesn't , it becomes (1-1)

Answer 4

ok but then why does cosx become 1-1

Answer 5

okay do you know that \[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\] or not??

Answer 6

yes i do get that

Answer 7

because cos0 =1 and 1-1=0

Answer 8

ohhhh ok then..... but lets say the bottom of the x under the sinx was squared would it still give me the same answer

Answer 9

let's see about that ..
\[\frac{ sinx }{ x }*\frac{ (1-cosx)(1+cosx) }{x(1+cosx) }\]

Answer 10

\[\frac{ sinx }{ x }\frac{ \sin^2(x) }{ x(cosx+1) }=\frac{ \sin^2(x) }{ x^2 }*\frac{ sinx }{ cosx+1 }\]

Answer 11

\[=1^2*\frac{ 0 }{ 2 }=0\]

Answer 12

since both of them gave me zero can it be done by just placing the \[x^{2}\] underneath the \[\sin ^{2}x\] and solving it the same way you did the first time

Answer 13

yes , actually \[\lim_{x \rightarrow 0} \frac{ 1-cosx }{ x }=0\] is a rule , but I thought it's better to show you how it comes

Answer 14

ok thanks

Answer 15

welcome