Hello,
I have a problem solving a limit of a number series which is given with the n-th term:
dome123,
this first post has n approaching negative infinity. It should have been n approaching infinity, from the start.
You could let \(\displaystyle \ n \ = \ x^2\).
\(\displaystyle \displaystyle\lim_{n \to \ \infty} \bigg(\sqrt{x^2 \ + \ \sqrt{x^2 \ + \ x \ }} \ - \ x\bigg) \)
Multiply by the conjugate in each place, similarly in post #5, to get
\(\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{\sqrt{x^2 \ + \ x}}{\sqrt{x^2 \ + \ \sqrt{x^2 \ + \ x \ }} \ + \ x} \)
This will reduce the number of square root signs from six in post #4 down to three.
\(\displaystyle \sqrt{x^2} \ = \ |x|, \ \ \) but you can ignore that and write \(\displaystyle \ \sqrt{x^2} \ = \ x \ \ \)as you factor it out from the numerator and
the denominator, because \(\displaystyle \ x \to \ \infty \).
\(\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{\sqrt{x^2( 1 \ + \ \tfrac{1}{x})}}{\sqrt{x^2 \ + \ \sqrt{x^2( 1 \ + \ \tfrac{1}{x})} } \ + \ x} \)
\(\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{x \sqrt{ 1 \ + \ \tfrac{1}{x}}}{\sqrt{x^2 \ + \ x\sqrt{ 1 \ + \ \tfrac{1}{x}} } \ + \ x} \)
\(\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{x \sqrt{ 1 \ + \ \tfrac{1}{x}}}{\sqrt{x^2( 1 \ + \ \tfrac{1}{x}\sqrt{ 1 \ + \ \tfrac{1}{x}}) } \ + \ x} \)
\(\displaystyle \displaystyle\lim_{n \to \ \infty} \dfrac{x \sqrt{ 1 \ + \ \tfrac{1}{x}}}{x\sqrt{ 1 \ + \ \tfrac{1}{x}\sqrt{ 1 \ + \ \tfrac{1}{x}} } \ + \ x} \)
Continue factoring out the x variables, divide them/cancel them out, and continue.