- #1

- 1

- 0

x->0

My attempt:

lim [1 - cos (x)/ x][lim 1/x] = (1)(1/0)

Please help me with some basic examples.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Halcyon99
- Start date

- #1

- 1

- 0

x->0

My attempt:

lim [1 - cos (x)/ x][lim 1/x] = (1)(1/0)

Please help me with some basic examples.

- #2

- 176

- 0

lim [1 - cos (x)/ x]=1?

How ??

While solving limit Problems, you make use of certain solutions

like

[tex]\lim_{x\rightarrow 0} \frac{Sin(x)}{x}=1[/tex]

[tex]\lim_{x\rightarrow 0} \frac{e^x-1}{x}=1[/tex]

==============================================

eg:[tex]\lim_{x\rightarrow 0} \frac{Tan(x)}{x}[/tex]

=>[tex]\lim_{x\rightarrow 0} \frac{\frac{Sin(x)}{cosx}}{x}[/tex]

==>[tex]\lim_{x\rightarrow 0} \frac{Sin(x)}{x} \lim_{x\rightarrow 0} \frac{1}{Cosx}[/tex]

==>1*1=1

end example==================

In this case,try to convert this into a known limit

HINT:[tex]cos2x=1-2sin^2x[/tex]

How ??

While solving limit Problems, you make use of certain solutions

like

[tex]\lim_{x\rightarrow 0} \frac{Sin(x)}{x}=1[/tex]

[tex]\lim_{x\rightarrow 0} \frac{e^x-1}{x}=1[/tex]

==============================================

eg:[tex]\lim_{x\rightarrow 0} \frac{Tan(x)}{x}[/tex]

=>[tex]\lim_{x\rightarrow 0} \frac{\frac{Sin(x)}{cosx}}{x}[/tex]

==>[tex]\lim_{x\rightarrow 0} \frac{Sin(x)}{x} \lim_{x\rightarrow 0} \frac{1}{Cosx}[/tex]

==>1*1=1

end example==================

In this case,try to convert this into a known limit

HINT:[tex]cos2x=1-2sin^2x[/tex]

Last edited:

- #3

- 621

- 1

use le'hopitals(spelling?) rule

Last edited:

- #4

berkeman

Mentor

- 60,829

- 11,216

poolwin2001 said:lim [1 - cos (x)/ x]=1?

Hmm, I got 1/2. I think you left the x^2 out of the original denominator:

lim x-->0 of ( 1-cos(x) )/x^2

= lim x-->0 ( sin(x) )/2x

= lim x-->0 ( cos(x) )/2 = 1/2

BTW, here's a good page on L'Hopital's Rule:

http://www.math.hmc.edu/calculus/tutorials/lhopital/

PS -- I went back and used a calculator to plug in small numbers for x, and it looks more like the limit of the original function of x goes to zero. I wonder if I did the first differentiation wrong, or the calculator is fooling me. What do other folks get?

Last edited:

- #5

- 1

- 0

I guess this can be solved in eitherof following 2 ways.:

1. Using formula of cosx=1-(x^2)/2!+(x^4)/4!...

so that

(1-cosx)=x^/2-(x^4)/4!....

=x^2[1-(x^2)/4!....]

so that

Lt x->0 of (1-cosx)/x^2 becomes :

=Lt x->0 of 1/2[1-(x^2)/4!....]

=1/2

2. other method could be usin L'Hospital's rule using differentiation

this too yields answer as 1/2

take care

- #6

- 695

- 0

PS -- I went back and used a calculator to plug in small numbers for x, and it looks more like the limit of the original function of x goes to zero. I wonder if I did the first differentiation wrong, or the calculator is fooling me. What do other folks get?

Your calculator was probably using degrees instead of radians.

- #7

berkeman

Mentor

- 60,829

- 11,216

Perfect! Thanks, Muzza. I thought I was losing my mind!

I was going to suggest to Halcyon99 that a good way to check the results you get from L'Hospital's Rule is to just plug some small numbers into the original fractional expression using a calculator. Now I know to add the caveat that the calculator needs to be in Radian mode. I'd spaced the part about when series expansions or other expressions are shown sharing a variable between trig functions and algebraic functions, the trig function arguments are in radians.

Thanks again, -Mike-

Share: