Question

find the derivative using the definition of limits.
The function is as follows.

Answer 1

\[f(x) = \sin (x)\]

Answer 2

I know that I am supposed to use

\[\lim_{h \rightarrow 0} { {\sin(x+h) - \sin(x) } \over {h}}\]

Answer 3

Use the series expansion of sin, it makes it easier.

Answer 4

\[\lim_{h \rightarrow 0}\frac{\sin(x+h)-sinx}{h}=\lim_{h \rightarrow 0}\frac{sinxcosh+sinhcosx-sinx}{h}\]
\[=\lim_{h \rightarrow 0}(\frac{sinx(\cosh-1)}{h}+\frac{sinhcosx}{h})\]

Answer 5

(cosh-1)/h->0
sinh/h->1
as h->0

Answer 6

if you knew the series expansion you would already know the derivative!

Answer 7

Hello guru :)

Answer 8

wow slow again. hello to you my guru.

Answer 9

\[sinx \lim_{h \rightarrow 0}\frac{\cosh-1}{h}+cosx \lim_{h \rightarrow 0}\frac{\sinh}{h}=sinx(0)+cosx(1)=0+cosx=cosx\]

Answer 10

Sorry I meant the exponential definition

Answer 11

i think the way i describe is really easy

Answer 12

:) im no more a guru now

Answer 13

hes a horse's rear get it straight satellite

Answer 14

could you explain me why \[\lim_{h \rightarrow 0}{ {\sin(h) } \over {h}} = 1\]

Answer 15

I cannot use L'Hospitals rule yet

Answer 16

\[\lim_{h\rightarrow 0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\rightarrow 0}\frac{\sin(x)\cos(h)+\cos(x)sin(h)-\sin(x)}{h}\]
\[=\lim_{h\rightarrow 0}\frac{\sin(x)(\cos(h)-1)}{h}+\lim_{h \rightarrow 0}\frac{\cos(x)\sin(h)}{h}\]
\[=\sin(x) \lim_{h \rightarrow 0}\frac{\cos(h)-1}{h}+\cos(x)\lim_{h \rightarrow 0}\frac{\sin(h)}{h}\]
\[=\sin(x)\times 0+\cos(x) \times 1 = \cos(x)\]

Answer 17

that is why they prove that limit in the text by something called the "squeeze theorem"

Answer 18

but if you use the squeeze theorem,

-1/h -> - infinity 1/h -> infinity
so we cannot "squeeze" sin(h) / h

Answer 19

by squeeze thm, sinh/h->1 as h->0