Evaluating Limit: [(1/x+4)-1/4]/x

[Johnnycab]

Im asked to evaluate:

[(1/x+4)-1/4]/x
Lim -> 0

Substitution, factoring and conjugate multiplication don't work

The question tells me to multipy the top and bottam by the lcd of the little fractions namely 4(x+4)

--here I am a little confused, did all the book do is take both denominators (x+4) and 4 and make them the LCD together?

So like if my question was for example like this,

[(1/2x+3)-1/4]/x

My LCD would be 4(2x+3), am i wrong?

Thanks

 

[TMFKAN64]

No, that's right. To be honest, you are OK just putting them over a common denominator... it doesn't strictly speaking have to be the LCD.

That is, if you had:

[(1/2x+4)-1/4]/x

Your LCD would be 4(x+2), but you could solve the problem just fine if you multiplied the numerator and denominator by 4(2x+4).

 

[Johnnycab]

thanks

ok i understand multiplying the numerator and denominator by 4(x+4),and to find a LCD but the next steps are just a little confusing

lim x->0
(1)[(1/x+4)-1/4]/x

(2) [(1/x+4)-1/4]/x
x
4(x+4)
4(x+4)
=
(3) [4-(x-4)]/[4x(+4)]
=
(4) -x/[4x(x+4)]
=
(5) -1/[4x(x4)]

Now i can substitute

Im sorry if this doesn't make sense but I am not understanding this way of solving limits. I broke it into the 5 steps, can this way be described in a step process?

Thanks

 

[TMFKAN64]

Generally, you seem to be doing the right thing, although it's hard to see because of your insistance on using "x" for both the variable and the multiplication operator!

But yes, the general idea is you do some algebraic manipulations (in as many steps as you need!) until you get to an expression that you know how to take the limit of directly.

 

[HallsofIvy]

By [(1/x+4)-1/4]/x
Lim -> 0

Do you mean [1/(x+4)- 1/4]/x or do you mean [(1/x)+ 4- 1/4]/x. What you wrote should be strictly interpreted as the latter but that does not converge as x goes to 0. The former is the same as
\frac{\frac{4}{4(x+4)}-\frac{x+4}{4(x+4)}}{x}= \frac{-1}{4(x+4)}
and the limit of that should be easy.

 

[Johnnycab]

By [(1/x+4)-1/4]/x
Lim -> 0

Do you mean [1/(x+4)- 1/4]/x or do you mean [(1/x)+ 4- 1/4]/x. What you wrote should be strictly interpreted as the latter but that does not converge as x goes to 0. The former is the same as
\frac{\frac{4}{4(x+4)}-\frac{x+4}{4(x+4)}}{x}= \frac{-1}{4(x+4)}
and the limit of that should be easy.

Yes i mean [1/(x+4)- 1/4]/x

I see this equation solved out, but I am not understanding it

Like after i find out that the LCD of the simple fractions in the numerator is[4(x-4)], what happened at steps 3 to 5?