Limes of a function

[mathwannabe]

Please, help me find a

\(\displaystyle \lim_{x \rightarrow 0}\dfrac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}\)

BUT, without the use of L'Hospitals rule, only by using algebraic manipulation. No matter what kind of insane rationalizations I try, I always end up with 0/0... The book provides the answer 3/2.

I have just managed to do it like this:

Using the rule that \(\displaystyle \lim_{x \rightarrow 0}\dfrac{\sqrt[n]{1+x}-1}{x}=\dfrac{1}{n}\)

So I just divided both the numerator and the denominator by \(\displaystyle x\) and then took the limes (ok, ok, limit xD). Limes sound to sour anyway

 

[JeffM]

Please, help me find a

\(\displaystyle \lim_{x \rightarrow 0}\dfrac{\sqrt{1+x}-1}{\sqrt[3]{1+x}-1}\)

BUT, without the use of L'Hospitals rule, only by using algebraic manipulation. No matter what kind of insane rationalizations I try, I always end up with 0/0... The book provides the answer 3/2.

Hey there mathwannabe. How are things going? Happy about going back to school?

\(\displaystyle u = \sqrt[6]{1 + x} \implies \dfrac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} = \dfrac{u^3 - 1}{u^2 - 1} = \dfrac{(u^2 + u + 1)(u - 1)}{(u + 1)(u - 1)} =\)

\(\displaystyle \dfrac{u^2 + u + 1}{u + 1} = \dfrac{u(u + 1) + 1}{u + 1} = u + \dfrac{1}{u + 1} = \sqrt[6]{x + 1} + \dfrac{1}{\sqrt[6]{x + 1} + 1}.\)

Now take your limit.

If I remember correctly, your very first post, back when you were prepping for your entrance exam, involved difference of powers. You forgot it then. You forgot it now.

You have posted very seldom so I presume most of your studies have gone well.

Edit: By the way, "limes" means boundary or border in Latin, but in English "limes" refers to a kind of fruit like a lemon, but smaller with a green skin. The word you want in English is "limit."

 

[Deleted member 4993]

Hey there mathwannabe. How are things going? Happy about going back to school?

\(\displaystyle u = \sqrt[6]{1 + x} \implies \dfrac{\sqrt{1 + x} - 1}{\sqrt[3]{1 + x} - 1} = \dfrac{u^3 - 1}{u^2 - 1} = \dfrac{(u^2 + u + 1)(u - 1)}{(u + 1)(u - 1)} =\)

\(\displaystyle \dfrac{u^2 + u + 1}{u + 1} \)

Now take your limit u → 1.
.

Made it a bit shorter

 

[JeffM]

Made it a bit shorter

Made it quite a bit shorter

 

[mathwannabe]

Thank you both Yes, the studies are going fine, math logic, English language, programming 1, linear algebra... no problem there. Mathematical analysis is a bit trickier though.

Now I want to cover myself with my ears for not seeing that solution...