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

• Johnnycab
In summary, at step 3 you substituted the LCD of the little fractions (4(x-4)) into the equation to get a result that did not converge. At step 4 you divided the numerator (x+4) by the denominator (4(x-4)) which yielded a negative result. At step 5 you substituted the LCD of the little fractions (4(x-4)) into the equation to get a result that did not converge.
Johnnycab

[(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

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.

[(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).

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

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.

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.

HallsofIvy said:
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?

1. What is the limit of [(1/x+4)-1/4]/x as x approaches infinity?

The limit of [(1/x+4)-1/4]/x as x approaches infinity is 0. This can be determined by simplifying the expression and applying the limit rules for rational functions.

2. How do you evaluate the limit of [(1/x+4)-1/4]/x as x approaches 0?

To evaluate the limit of [(1/x+4)-1/4]/x as x approaches 0, first simplify the expression to (1+4x)/4x^2. Then, apply the limit rules for rational functions to get the answer of 1/4.

3. Can the limit of [(1/x+4)-1/4]/x be evaluated using direct substitution?

No, the limit of [(1/x+4)-1/4]/x cannot be evaluated using direct substitution because it results in an undefined expression (0/0) when x is substituted in. Instead, other methods such as simplification and applying limit rules must be used.

4. Is the limit of [(1/x+4)-1/4]/x continuous at x = 0?

Yes, the limit of [(1/x+4)-1/4]/x is continuous at x = 0 because the left-hand and right-hand limits are equal to 1/4 at this point.

5. What is the difference between evaluating a limit and finding the limit?

Evaluating a limit involves finding the numerical value of the limit at a specific point or as x approaches a specific value. Finding the limit involves determining if a limit exists and what the value of the limit is at a certain point or as x approaches infinity or negative infinity.

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