limit help

[khryss]

lim (sqrt(x+5)-sqrt(5))/X
x->0

tried to multiply by the conjugate, but got 1/0 and I get the right answer with l'hospitals rule but I cant use it for this question

 

[MarkFL]

Rationalizing the numerator is the way to go, and you should not get 1/0, you should find a determinate form. Can you post your work so we can see where you've gone wrong in the process?

 

[khryss]

multiplying by conjugate
numerator: sqrt(x+5)-sqrt(5) * sqrt(x+5)+sqrt(5)
=>> (x+5)-5 => x

denominator: x * sqrt(x+5)[+]sqrt(5) ==>

oops I just realized that on paper I subtracted in the denominator instead of adding... thanks! I got the right answer now

= 1/2sqrt(5)

 

[MarkFL]

Good deal!

I do recommend using brackets to make your expressions more clear, for instance I would write the result, in lieu of \(\displaystyle \LaTeX\), as:

1/(2sqrt(5))

this way it is clear that you mean \(\displaystyle \dfrac{1}{2\sqrt{5}}\) and not \(\displaystyle \dfrac{1}{2}\sqrt{5}\).

 

[Bob Brown MSEE]

Taylor Series of numerator is:
\(\displaystyle x/(2 Sqrt[5]) - x^2/(40 Sqrt[5]) + x^3/(400 Sqrt[5]) - x^4/(
3200 Sqrt[5])\) ...
(like l'hospitals rule w/o using l'hospitals rule)

 

[MarkFL]

Taylor Series of numerator is:
\(\displaystyle x/(2 Sqrt[5]) - x^2/(40 Sqrt[5]) + x^3/(400 Sqrt[5]) - x^4/(
3200 Sqrt[5])\) ...
(like l'hospitals rule w/o using l'hospitals rule)

I recommend the \(\displaystyle \LaTeX\) code:

\dfrac{x}{2\sqrt{5}}-\dfrac{x^2}{40\sqrt{5}}+\dfrac{x^3}{400\sqrt{5}}-\dfrac{x^4}{3200\sqrt{5}}+\cdots

to get:

\(\displaystyle \dfrac{x}{2\sqrt{5}}-\dfrac{x^2}{40\sqrt{5}}+\dfrac{x^3}{400\sqrt{5}}-\dfrac{x^4}{3200\sqrt{5}}+\cdots\)

 

[Bob Brown MSEE]

Recommend \(\displaystyle \LaTeX\) code:

Thanks!