L'Hopital's Rule

[chelle2007]

Hello! I'm not sure if I am doing this correctly or not.

Evaluate the limit using L'Hopital's Rule:

lim (as x approaches 0+) square root of x ((ln(x))

I made it into a quotient so it was ln(x) / (1/square root of x)

Is this correct??

 

[soroban]

Hello, chelle2007!

Evaluate the limit using L'Hopital's Rule:

. . \(\displaystyle \displaystyle \lim_{x\to 0^+}\sqrt{x}\cdot\ln x\)

I made it into a quotient so it was: .\(\displaystyle \dfrac{\ln x}{\frac{1}{\sqrt{x}}} \)

Is this correct?

It is correct, but I would write it: .\(\displaystyle x^{\frac{1}{2}}\cdot\ln x \:=\:\dfrac{\ln x}{x^{-\frac{1}{2}}} \)

 

[medicalphysicsguy]

Why a quotient?

Hello! I'm not sure if I am doing this correctly or not.

Evaluate the limit using L'Hopital's Rule:

lim (as x approaches 0+) square root of x ((ln(x))

I made it into a quotient so it was ln(x) / (1/square root of x)

Is this correct??

May I ask why you made it into a quotient? Doesn't it make it more difficult to put x in the denominator when you are approaching zero? Isn't the product version easier to evaluate?

I am working on L'Hopital's right now too so I am just curious.

Thanks!

 

[soroban]

Hello, medicalphysicsguy!

May I ask why you made it into a quotient?
Doesn't it make it more difficult to put x in the denominator when you are approaching zero?
Isn't the product version easier to evaluate?

L'Hopital's Rule requires a quotient.


\(\displaystyle \displaystyle\text{If }\lim_{x\to a} \frac{f(x)}{g(x)}\,\text{ produces an indeterminate form,}\)
. . \(\displaystyle \displaystyle\text{we can evaluate: }\lim_{x\to a} \frac{f'(x)}{g'(x)}\)