Solve the limit question

[Jan Hill]

Homework Statement

Find the limit as x--> 0 of numerator = 1/x-1 + 1/x + 1 denominator = x

2.

3.
I multiplied the numerator by the conjugate and therefore the denominator by the conjugate but in simplifying, I still get zero in the denominator and this will not do. Am I going about it in the wrong way?

The Attempt at a Solution

 

[lanedance]

do you mean
$$\frac{1/x-1 + 1/x + 1}{x}$$

i would start by multiplying thrrugh x/x & (x-1)/(x-1) to simplify the numerator

 

[Jan Hill]

No the question is

Find the limit as x--> 0 of 1/x -1 + 1/x + 1 all divided by x

 

[HallsofIvy]

That is exactly what lanedance wrote and you said was wrong!

Please use parentheses! You probably mean "1/(x- 1)+ 1/(x+ 1)" all divided by x.

That is
$$\frac{\frac{1}{x-1}+ \frac{1}{x+ 1}}{x}$$
Go ahead and add the fractions in the numerator. What is a common denominator and what do you get when you add the fractions?

 

[Mark44]

Find the limit as x--> 0 of numerator = 1/x-1 + 1/x + 1 denominator = x

Please start learning how to format mathematical expressions properly, instead of "numerator = ..." and "denominator = ..."

At the very least, you can write the expression above in text as
[1/(x - 1) + 1/(x + 1)]/x

What you wrote could reasonably be interpreted as [(1/x) - 1 + (1/x) + 1]/x, but I don't think that's what you meant.

If you want to get fancier, you can see how lanedance formatted the LaTeX expression he wrote by clicking that expression.