Is there another way to solve this limit: lim(x→0+) of (1/x +2)^1/2 - (1/x)^1/2

[hndalama]

lim(x→0+) of (1/x +2)^1/2 - (1/x)^1/2

I solve the limit by expressing the equation with a common denominator, root x. Then I use L'Hopital's rule to find the limit. The problem is that this question is given before L'Hopital's rule has been introduced in the course, I just happened to know about it from a previous course. So I am asking if there is another way of solving this limit.

 

[pka]

lim(x→0+) of (1/x +2)^1/2 - (1/x)^1/2

I solve the limit by expressing the equation with a common denominator, root x. Then I use L'Hopital's rule to find the limit. The problem is that this question is given before L'Hopital's rule has been introduced in the course, I just happened to know about it from a previous course. So I am asking if there is another way of solving this limit.

You do mean \(\displaystyle \left[ {{{\left( {\sqrt {x + 2} } \right)}^{ - 1}} - {{\left( {\sqrt x } \right)}^{ - 1}}} \right]\)? Do you not?

 

[hndalama]

You do mean \(\displaystyle \left[ {{{\left( {\sqrt {x + 2} } \right)}^{ - 1}} - {{\left( {\sqrt x } \right)}^{ - 1}}} \right]\)? Do you not?

No not quite, I mean sqrt[(1/x) + 2] - sqrt[1/x]

 

[pka]

No not quite, I mean sqrt[(1/x) + 2] - sqrt[1/x]

Limit as \(\displaystyle x\to 0^+~: \)
\(\displaystyle \begin{align*} \sqrt{x^{-1}+2}-\sqrt{x^{-1}}&=\dfrac{\sqrt{1+2x}-1}{\sqrt x} \\&=\dfrac{2x}{\sqrt x(\sqrt{1+2x}+1)}\\&=\dfrac{2\sqrt x}{\sqrt{1+2x}+1}\\&\to 0 \end{align*}\)

 

[hndalama]

Limit as \(\displaystyle x\to 0^+~: \)
\(\displaystyle \begin{align*} \sqrt{x^{-1}+2}-\sqrt{x^{-1}}&=\dfrac{\sqrt{1+2x}-1}{\sqrt x} \\&=\dfrac{2x}{\sqrt x(\sqrt{1+2x}+1)}\\&=\dfrac{2\sqrt x}{\sqrt{1+2x}+1}\\&\to 0 \end{align*}\)

thank you