How can I simplify this equation and see the connection between #5 and #7?

[SyntheticVisions]

Technically this is a calculus problem I'm working on, but I'm just having problems with the Algebra portion.

If I have:

(\frac{1}{x\sqrt{1+x}} - \frac{1}{x})

How can I simply this so that I can substitute in 0 for x?

 

[robert Ihnot]

You want to get this in a form for the use of L'Hospital's Rule: \frac{1-\sqrt{1+x}}{x(\sqrt{1+x})}

In this form we see that as x\rightarrow0 the quotient is undefined, so we can differentiate and simplify.

 

[SyntheticVisions]

We haven't gone into differentiation or anything like that, is there another way?

Actually, the problem that I'm trying to figure out is

lim
x -> 0 of the expression above.


edit: For clarification - it's not for homework, it's just a problem I'm trying to figure out.

 

[robert Ihnot]

I don't know any other way to do this problem. This is how you do it using the Calculus. You differentiate and get:

\frac{[-2\sqrt{1+x}]^-1}{(2+3x)[2\sqrt{1+x}]^-1}=\frac{-1}{2+3x}\rightarrow \frac{-1}{2} ...as... x \rightarrow 0

 

[Fredrik]

The substitution

t=\sqrt{1+x}

simplifies the function to

-\frac{1}{(1+t)t}

The limit of this as t goes to 1 is -1/2.

 

[robert Ihnot]

The substitution

t=\sqrt{1+x}

simplifies the function to

-\frac{1}{(1+t)t}

The limit of this as t goes to 1 is -1/2.

That looks like a better way!

 

[shmoe]

To add yet another way, rationalize the numerator. Multiply

\frac{1-\sqrt{1+x}}{x(\sqrt{1+x})}

by

\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}

to get

\frac{-x}{x(\sqrt{1+x})(1+\sqrt{1+x})}

and go from there.

 

[Gokul43201]

And finally see that #5 and #7 are really doing the same thing.

They's both making use of the fact that x can be factored as -(1-\sqrt{1+x})(1+\sqrt{1+x})