Question

Use calculator to find limits: when x approaches 0 for Lim (1-cosx)/x ;
when x approaches infinity for Lim x/square root (3x^2-4)

Answer 1

do u know the basic limit identity of limit x->0 sin x/x = ???

Answer 2

or the question is to just use calcy and not do it analytically??

Answer 3

the question asks to use calculator and give 2 decimal places

Answer 4

still,u need to know limit x->0 sin x/x = ?
do u know the value of this limit?

Answer 5

umm, I don't know

Answer 6

ok,\[\lim_{x \rightarrow 0}\frac{\sin x}{x}=1\]
we are going to use this formula.
first multiply (1+cos x) in numerator and denominator and tell me what u get in numerator.,after simplification??

Answer 7

I got sinx/(x+cosx^2)

Answer 8

seems incorrect,
there was (1-cos x) in the numerator,i multiplied (1+cos x)
so(1-cos x)(1+cos x)=1-cos^2 x = sin^2 x in the numerator....
did u get this??
there was x in denominator,i multiplied (1+cos x) to just get x(1+cos x) in denominator...

Answer 9

yeah, I get it right, but I miss typed the sin^2 x in the numerator here.

Answer 10

but u also got denominator as (x+cos^2 x)
but actually,it is x(1+cos x)
anyways,going ahead,i will split sin^2 x as sin x * sin x and write the limit as
\[\lim_{x \rightarrow 0}\frac{\sin x}{x}\frac{\sin x}{1+\cos x}\]
did u get this step??this is most important step.........

Answer 11

but u also got denominator as (x+cos^2 x)
but actually,it is x(1+cos x)
anyways,going ahead,i will split sin^2 x as sin x * sin x and write the limit as
\[\lim_{x \rightarrow 0}\frac{\sin x}{x}\frac{\sin x}{1+\cos x}\]
did u get this step??this is most important step.........

Answer 12

yeah, I get this one

Answer 13

so now i will split the limits like this :
\[\lim_{x \rightarrow 0}\frac{\sin x}{x}\lim_{x \rightarrow 0}\frac{\sin x}{1+\cos x}\]
we know the value of 1st limit will be 1
the value of 2nd limit is evaluated by just putting x=0,u know the value of sin 0 and cos 0 ??

Answer 14

So sin0=0 and cos0=1?

Answer 15

yup correct,so just substitute that and tell me what u get the value of 2nd limit??

Answer 16

Is it 0?

Answer 17

yes it is :) all this work to get the final answer as 0!!
correct to 2 decimal place,its 0.00 :P
but u got all the steps?ask if u still have doubts anywhere...

Answer 18

haha, that's ironic... it's a heavy process, but I get it now.

Answer 19

hint for 2nd question,divide x in numerator and denominator....

Answer 20

ask if u have difficulty solving the 2nd.....