limits also on with trig

brodie6 · **Posted:** Sat, 2 Oct 2010 09:26:13 UTC

lim x->0 [sqrt(x+9)-3] /x goes to 1/6 and i have no clue how

lim x->0 (x-sinx)/x^3 also goes to 1/6 and i have no clue how really need

help im taking enriched calculus 1

brodie6 · **Posted:** Sat, 2 Oct 2010 09:29:36 UTC

well i mean i think ive come close but cant quite get it to make sense

outermeasure · **Posted:** Sat, 2 Oct 2010 10:45:02 UTC

brodie6 wrote:

lim x->0 [sqrt(x+9)-3] /x goes to 1/6 and i have no clue how

lim x->0 (x-sinx)/x^3 also goes to 1/6 and i have no clue how really need

help im taking enriched calculus 1

Please show your work.

Hint for the first question: $\dfrac{\sqrt{x+9}-3}{x}=\dfrac{\sqrt{x+9}-3}{x}\cdot\dfrac{\sqrt{x+9}+3}{\sqrt{x+9}+3}$

Hint for the seond question: Apply (one of the many) definitions of sin(x).

brodie6 · **Posted:** Sat, 2 Oct 2010 18:27:51 UTC

for the first one that was deffinatly my first move and it all makes sense now it was just late and i was missing that the x s cross once you get 9-9 = 0 and the bottom goes to 6 when 3+3 i am thinking of this correctly right ? , for the trig one i dont understand any of the notation your using ive tried setting sinx to sqrt(1-cox^2) which i think is close since i think it will have to be a 1+cosx twice on the bottom or ive tried the conjugate top and bottom by x+sinx to deal with sinx^2 instead but i just run into zeros this way as well and just feels like im going in circles once i get x- sqrt(1-cosx^2) am i allowed to go conjugate sqrt(1+cosx^2) then use the ratio of the x/x^3 to get 1/3 then go 3(2) is this the right track

outermeasure · **Posted:** Sun, 3 Oct 2010 05:44:13 UTC

Which rigorous definition of sin(x) are you using?

brodie6 · **Posted:** Sun, 3 Oct 2010 06:08:39 UTC

i cant tell if your being sarcastic about the rigorous deffinition but ive been trying sqrt(1-cox^2) and sqrt(1-cos2x/2) is this not deep enough i must be a fool cuz it has me losing sleep and i can do mostly every other limit problem in my book or i keep doing the same mistake over and over

outermeasure · **Posted:** Sun, 3 Oct 2010 06:25:41 UTC

brodie6 wrote:

i cant tell if your being sarcastic about the rigorous deffinition but ive been trying sqrt(1-cox^2) and sqrt(1-cos2x/2) is this not deep enough i must be a fool cuz it has me losing sleep and i can do mostly every other limit problem in my book or i keep doing the same mistake over and over

One definition of $\sin(x)$ is $\displaystyle\sin(x):=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$ for all $x\in\mathbb{C}$ . See what you can get from here.

(There are other equivalent definitions of the sine, such as using Euler's sine product or uniqueness to solutions of differential equations. But to say "sin(x) is the y-coordinates of ... when the angle is x", without a definition of angle that doesn't make the definition circular, is not a definition of sin(x).)

alstat · **Posted:** Mon, 4 Oct 2010 05:09:24 UTC

Outermeasure's just trying to psych you out - you won't see that definition of sine for a long time, since the problem you are doing is first semester Calculus.

Just split up $\frac{x-\sin x}{x^3}$ into two fractions:
$= \frac{x}{x^3} - \frac{\sin x}{x^3} = \frac{1}{x^2} - \frac{\sin x}{x}\frac{1}{x^2}$

Check your answer - you should be able to get each of the two resulting limits to be zero, especially since you should just have learned a limit theorem involving $\frac{\sin x}{x}$ . Shouldn't be $\frac{1}{6}$ .

outermeasure · **Posted:** Mon, 4 Oct 2010 05:23:16 UTC

alstat wrote:

Outermeasure's just trying to psych you out - you won't see that definition of sine for a long time, since the problem you are doing is first semester Calculus.

Just split up $\frac{x-\sin x}{x^3}$ into two fractions:
$= \frac{x}{x^3} - \frac{\sin x}{x^3} = \frac{1}{x^2} - \frac{\sin x}{x}\frac{1}{x^2}$

Check your answer - you should be able to get each of the two resulting limits to be zero, especially since you should just have learned a limit theorem involving $\frac{\sin x}{x}$ . Shouldn't be $\frac{1}{6}$ .

No! The limit is indeed 1/6. The poles parts cancel, but you need to consider the finite part of $\frac{\sin x}{x}\cdot\frac{1}{x^2}$ , namely the order x^2 part of $\frac{\sin x}{x}$ .

aswoods · **Posted:** Mon, 4 Oct 2010 12:11:31 UTC

For the second one, use L'HÃ´pital's rule three times in a row.

Outermeasure's suggestion was to write sin x as the infinite series $x-\frac1{3!}x^3+\frac1{5!}x^5-\frac1{7!}x^7+...$ , a method which works perfectly well, but is probably not covered in your course.

alstat · **Posted:** Tue, 5 Oct 2010 01:45:54 UTC

I was wrong! My apologies - never try this on a lack of sleep!
My question is, is L'Hopital's rule available in Calculus 1?
It usually is only about a chapter ahead of the infinte series that would also solve it. Usually both L'Hopital & series are in 2nd semester Calc.

Shadow · **Posted:** Tue, 5 Oct 2010 01:57:36 UTC

alstat wrote:

My question is, is L'Hopital's rule available in Calculus 1?

Depends on who's teaching it.

S.O.S. Mathematics CyberBoard

limits also on with trig

Who is online