Question

Evaluate the limit using l'Hopital's Rule
lim(x->0+)((1/x)-(1/sqrtx))

Answer 1

\[\lim(x->0+)(\frac{ 1 }{x }-\frac{ 1 }{ \sqrt{x} })\]

Answer 2

I tried rewriting it with an LCD like my teacher suggested, and then using l'hopital, but I got an indeterminate again after I did so

Answer 3

You need to write\[\left( \frac{ 1 }{ x } -\frac{ 1 }{ \sqrt{x} }\right)\] as\[\frac{ f \left( x \right) }{ g \left( x \right) }\]i.e write the given function with a common denominator.

Answer 4

Yes

Answer 5

I got \[\frac{ \sqrt{x}-x }{ x }\]

Answer 6

over xsqrtx

Answer 7

not x

Answer 8

Which yielded 0/0, then I l'hopitaled and got infinity/infinity, which is an issue :/

Answer 9

OK. Now use l'Hopital's Rule.\[\lim_{x \rightarrow 0^+}\frac{ f \prime \left( x \right) }{ g \prime \left( x \right) }\]

Answer 10

What is \[f \prime \left( \sqrt{x} - x\right)\]

Answer 11

Ok. I got \[\frac{ \frac{ 1 }{ 2 }x ^{\frac{ -1 }{ 2 }}-1 }{ x(\frac{ 1 }{ 2x ^{\frac{ -1 }{ 2 }} }}\]

Answer 12

Whoa thats not how I wanted it to look wtf

Answer 13

The denominator is x((1/2)x^-1/2)+sqrtx

Answer 14

You mean\[\frac{ \frac{ 1 }{ 2 } x^{-\frac{ 1 }{ 2 }} -1 }{ \frac{ 1 }{ 2 } x^{-\frac{ 1 }{ 2 }}+x^{\frac{ 1 }{ 2} }}\]

Answer 15

Exactly

Answer 16

Wait no, there's an x in front of the first term of the denominator

Answer 17

You're right. I got lost in the typesetting.

Answer 18

Ok

Answer 19

What happens if you multiply those quantities together in the first term of the denominator?

Answer 20

x/2sqrtx

Answer 21

sqrtx/2

Answer 22

Do you not get\[\frac{ \frac{ 1 }{ 2 } x^{-\frac{ 1 }{ 2 }} -1 }{ \frac{ 1 }{ 2 } x^\frac{ 1 }{ 2 } +x^\frac{ 1 }{ 2 }}\]

Answer 23

It didn't even cross my mind to simplify that first term

Answer 24

oh my

Answer 25

Yes, pathetic, I know T_T

Answer 26

rational exponents might help in finding derivatives, but they stink in computation

Answer 27

Factor out the common x^{1/2} in the denominator. Factor out an x^{1/2} in the numerator so they will cancel and see what you've got.

Answer 28

You can't factor out an x^1/2 in the numerator.

Answer 29

Theres a -1 there

Answer 30

\[\frac{1}{x}-\frac{1}{\sqrt{x}}=\frac{1}{x}-\frac{\sqrt{x}}{x}=\frac{1-\sqrt{x}}{x}\]

Answer 31

rational exponents are messing you up
work with the radicals
you get the answer using l'hopital or even without it, once it is in that form

Answer 32

I still don't see how to get the sqrtx out of the numerator :/

Answer 33

1/2sqrtx -1

Answer 34

not sure what you mean
it is in the numerator
there is nothing you can do about it

Answer 35

Osprey triple said to factor a sqrtx out of the numerator and out of the denominator

Answer 36

in the expression
\[\frac{1-\sqrt{x}}{x}\] if the limit as not totally obvious (which it is) take the derivative top and bottom get the answer instantly
especially since the derivative of the denominator is just 1

Answer 37

where did you get that expression from

Answer 38

quit fooling around with rational exponents

Answer 39

\[\frac{1}{x}-\frac{1}{\sqrt{x}}=\frac{1}{x}-\frac{1}{\sqrt{x}}\times \frac{\sqrt{x}}{\sqrt{x}}=\frac{1-\sqrt{x}}{x}\]

Answer 40

I found an LCD between them

Answer 41

yes, it is \(x\)!

Answer 42

Oh. I got xsqrtx as mine

Answer 43

the lcd of \(\sqrt{9}\) and \(9\) is \(9\)

Answer 44

Well then I should be able to get x as my denominator. Right now I would get (3/2)sqrtx