lim sin h/2h
h0

using squeezing theorem .

Question

lim        sin h/2h
h>0

using squeezing theorem .

hartnn · Answer

do u know basic limit identity ? 
$\huge\lim \limits_{h \rightarrow0}\frac{sin h}{h}=1$

hartnn · Answer

from your Q,
$\huge\lim \limits_{h \rightarrow0}\frac{sin h}{2h}=\huge \frac{1}{2}\lim \limits_{h \rightarrow0}\frac{sin h}{h}=\frac{1}{2}$
got that ?

anonymous · Answer

how it can be 1/2 ?

hartnn · Answer

the 2 in the denominator is factored out as (1/2) because constants can be taken out of limits.
what remains is a standard limit formula which equals 1.

anonymous · Answer

@hartnn he is asking by squeeze principle..............this is the general way

hartnn · Answer

ohh...using squeezing theorem...

anonymous · Answer

ya...

hartnn · Answer

then u need to know this fact : 
$ \cos x \lt \frac{\sin x}{x} \lt 1$

hartnn · Answer

then using squeeze theorem u can directly say that 
$\huge\lim \limits_{h \rightarrow0}\frac{sin h}{h}=1$

hartnn · Answer

or, if u want to include 2 from the beginning, write 
$\large\frac{\cos x}{2} \lt \frac{\sin x}{2x} \lt \frac{1}{2}$
then using squeeze, u can direct get your answer.

anonymous · Answer

thanks ! it's really useful for me .

hartnn · Answer

welcome ^_^

anonymous · Answer

lim    2-cos3x-cos4x/x
x>0
using squeezing theorem .

hartnn · Answer

u must be knowing 
$-1 \lt \cos x \lt 1$
right ?

hartnn · Answer

rearrange that to get $\frac{1-cos x}{x}$ in between