Calculus assignment

**CenturionMonkey** · April 20th 2008, 04:37 PM

I'm doing an assignment on trig derivatives, limits, and graphing. I have no clue how to do this question though...

the limit as x approaches 0 of:

(tanx - sinx)/( $x^2$ tanx)

ps. sorry if you don't understand my question. I don't understand how to use that math font beyond exponents.

**TheEmptySet** · April 20th 2008, 04:49 PM

Originally Posted by CenturionMonkey

I'm doing an assignment on trig derivatives, limits, and graphing. I have no clue how to do this question though...

the limit as x approaches 0 of:

(tanx - sinx)/( $x^2$ tanx)

ps. sorry if you don't understand my question. I don't understand how to use that math font beyond exponents.

mult the top and bottom by 1/tanx to get

$\frac{1-\frac{\sin(x)}{\tan(x)}}{x^2}=\frac{1-\cos(x)}{x^2}$

there are a few ways to take this limit as x goes to zero the cool way is to expand cos as its power series

$\frac{1-(1-\frac{x^2}{2!}+\frac{x^4}{4!}-...)}{x^2}=\frac{\frac{x^2}{2!}-\frac{x^4}{4!}+...}{x^2}=\frac{\frac{1}{2}-\frac{x^2}{4!}+...}{1}$

taking the limit as x goes to zero gives 1/2

or we could use L'hospitals rule

we have the zero over zero form taking the derivatives we get

$\frac{1-\cos(x)}{x^2}=\frac{\sin(x)}{2x}$

use L.H again

$\frac{1-\cos(x)}{x^2}=\frac{\sin(x)}{2x}=\frac{\cos(x)}{2}$

taking the lim as x goes to zero gives 1/2 again.

Good luck.

**CenturionMonkey** · April 20th 2008, 05:02 PM

i'm thankful for your time guys, but i'm sorry i have no clue what L'hospital's rule is. We were just given 4 formulas

The limit as x approaches zero of:
sinx=0
cosx=1
(sinx)/(x)=1
(cosx - 1)(x)=0

and we are expected to solve these with those formulas. We can also use the other formulas we learned last year. i.e. double-angle formulas, addition/subtraction formulas...etc.

Thanks again

**Mathstud28** · April 20th 2008, 05:07 PM

Originally Posted by CenturionMonkey

i'm thankful for your time guys, but i'm sorry i have no clue what L'hospital's rule is. We were just given 4 formulas

The limit as x approaches zero of:
sinx=0
cosx=1
(sinx)/(x)=1
(cosx - 1)(x)=0

and we are expected to solve these with those formulas. We can also use the other formulas we learned last year. i.e. double-angle formulas, addition/subtraction formulas...etc.

Thanks again

Look at this http://www.mathhelpforum.com/math-he...orial.html...I show how to use it

Thread: Calculus assignment

LinkBack

Thread Tools

Display

Calculus assignment

Similar Math Help Forum Discussions

Some calculus problems in the assignment of Calculus 1 course

first calculus assignment of the year

Just a Few Calculus Assignment Questions

Calculus Assignment - HELP!!!

Major Calculus Assignment Issue

Search Tags