Limit [x->0] [ |3x + 1| - |3x - 1| ] / 2x

[ku1005]

Hey guys , with the followign limit im really stuck ,im just not sure how to deal with the inequalities, my workings are below, any help appreciated as always.

lim x->0 for [|3x+1|-|3x-1|]/2x

so i had

1/2 lim x->0 sqrt((3x+1)^2)-sqrt((3x-1)^2) / x

and then i thought you could divide through by the highest power of x (ie x outside sqrt and x^2 inside sqrt and try and involve sin x/ x which tends towards 1 as x->0

but so far i have been unsucessful in doing this

cheers

 

[stapel]

How in the world were you planning on bringing [sin(x)]/x into play?!? Why not just work with the absolute values?

The arguments, 3x + 1 and 3x - 1, change sign away from zero. For x close enough to zero, you can take the bars off: |3x + 1| = 3x + 1 for x > -1/3, and |3x - 1| = 3x = 1 for x < 1/3.

Simplify, and take the limit.

Eliz.

 

[ku1005]

thanks....and sorry...as it says on the forum policy thing....you shold show some of your workings and thinkings....otherwise you get ppl on here who just want assignent q's answered and hence they nvr learn from there mistakes.....thats all i was trying to do.....

 

[stapel]

I'm not quarrelling with showing your work or reasoning. I'm asking what that reasoning was. There are no trig functions in sight, so how were you planning on using sines? I don't understand.

Eliz.

 

[galactus]

It looks like you're making it more ifficult than it needs to be.

\(\displaystyle \L\\\frac{1}{2}\lim_{x\to\0}\frac{|3x+1|-|3x-1|}{x}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{(3x+1)-(3x-1)}{x}\)

L'Hopital:

\(\displaystyle \L\\3\lim_{x\to\0}[\text{sign}(3x+1)]-3\lim_{x\to\0}[\text{sign}(3x-1)]\)

Now, can you finish?. Not much more to go.

 

[pka]

As x approaches 0, we can assume that \(\displaystyle x \in \left[ {\frac{{ - 1}}{3},\frac{1}{3}} \right]\).

So, \(\displaystyle \left| {3x + 1} \right| - \left| {3x - 1} \right| = \left( {3x + 1} \right) - \left( { - 3x + 1} \right) = 6x\).

Thus \(\displaystyle \lim _{x \to 0} \frac{{\left| {3x + 1} \right| - \left| {3x - 1} \right|}}{{2x}} = 3\).