- #1

- 5

- 0

If I have:

[tex](\frac{1}{x\sqrt{1+x}} - \frac{1}{x})[/tex]

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

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- Thread starter SyntheticVisions
- Start date

- #1

- 5

- 0

If I have:

[tex](\frac{1}{x\sqrt{1+x}} - \frac{1}{x})[/tex]

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

- #2

- 1,056

- 0

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

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

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

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- #3

- 5

- 0

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.

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.

Last edited:

- #4

- 1,056

- 0

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

[tex]\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[/tex]

[tex]\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[/tex]

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- #5

Fredrik

Staff Emeritus

Science Advisor

Gold Member

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[tex]t=\sqrt{1+x}[/tex]

simplifies the function to

[tex]-\frac{1}{(1+t)t}[/tex]

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

- #6

- 1,056

- 0

Fredrik said:

[tex]t=\sqrt{1+x}[/tex]

simplifies the function to

[tex]-\frac{1}{(1+t)t}[/tex]

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

That looks like a better way!

- #7

shmoe

Science Advisor

Homework Helper

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[tex]\frac{1-\sqrt{1+x}}{x(\sqrt{1+x})}[/tex]

by

[tex]\frac{1+\sqrt{1+x}}{1+\sqrt{1+x}}[/tex]

to get

[tex]\frac{-x}{x(\sqrt{1+x})(1+\sqrt{1+x})}[/tex]

and go from there.

- #8

Gokul43201

Staff Emeritus

Science Advisor

Gold Member

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

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