Question

lim (sinx/x) when x-->0 means x goes to zero

when we substitute zero, we get 0/0
so we have to use hopital's rule, that's mean diffrentiate,
so we get (x cosx-sinx)/x^2 when we substitute zero,

we get 0/0 again

the book says it equals one !!

do you have any explanation

did I make a mistake in any step?

thanx

Answer 1

hey lion
you do derivative of top/derivative of bottom

Answer 2

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

Answer 3

\[\lim_{x \rightarrow 0}\frac{\cos(x)}{1}\]

Answer 4

You don't differentiate the same way you would when using L'Hopital's rule. You used the quotient rule which is correct in most cases. However, when using L'Hopital's rule, you differentiate the top and bottom independently. So you will get\[\lim_{x \rightarrow 0} [(\sin x)/x]\]Now you differentiate sinx and x.\[\lim_{x \rightarrow 0} \cos x/1\]Substituting the limit you get\[\cos (0)=1\]

Answer 5

That awesome work guys
thank you so much
I got the idea completely