Question

Find this limit. "No L'hopital's rule"

Answer 1

\[\large \lim_{h \rightarrow 0} \frac{\sin(x+h)- sinx}{h}\]

Answer 2

This is difference quotient for sin(x+H). Since the derivative of sin(x) is cos(x) SO PROCEED f-->>>

Answer 3

what I did was

\[\lim_{h \to 0} \frac{sinxcosh + cosxsinh - sinx}{h}\]

Answer 4

okay, then what tickles u...

Answer 5

Could I use that or do I have to use the difference quotient.

Answer 6

Well, I think @Psymon is going to tell you how to do it.

Answer 7

Sum of sines formula probably works:
sin(a + b) = sinacosb + sinbcosa

\[\frac{ sinxcosh + sinhcosx - sinx }{ h }\]
Group the two sinxterms so I can factor it out:
\[\frac{ sinxcosh-sinx + sinhcosx }{ h }= \frac{ sinx(cosx-1) + sinhcosx }{ h }\]
Split into two fractions:
\[\frac{ \sinx(cosh-1) }{ h }+\frac{ sinhcosx }{ h }\]Then take advantage of two of the limit identities:
\[\lim_{x \rightarrow 0}\frac{ sinx }{ x }= 1\]and
\[\lim_{x \rightarrow 0}\frac{ cosx-1 }{ x }= 0 \]
Which means the entire left fraction will becomes 0 since (cosh-1)/h becomes 0 as h goes to 0, also taking out the multiplied sinin the process. The right term becomes 1 * cos(0) since sinh/h is 1 as h goes to 0 by identity. And of course cos(0) is also just one, making the whole limit simply 1.

Answer 8

cool alright I can try this