evaluating limit using binomial series

Question

I am trying to evaluate the following limit by using binomial series $(1+x)^{1\over 2}=\left ({1/2}\over n\right) x^n$

$$\lim_{x\to 0}{{(1+x)^{1\over 2}-1-{1\over 2}x +{1\over 8}x^2}\over {x^3+x^5}}$$

I tried various manipulation and also used Maple to solve, and I found that this limit is the exact value of the binomial of 1/2 and 3, which is 0.0625, but I still couldn't see how you can manipulate this equation for the solution by hand.

Answer 1

Using the generalized binomial coefficients (here), Taylor expansion is $$\sqrt{1+x}=\sum_{n=0}^\infty\binom{\frac{1}{2}}{n} x^n=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5 x^4}{128}+\frac{7 x^5}{256}+O\left(x^6\right)$$ So, the numerator is $$\sum_{n=3}^\infty\binom{\frac{1}{2}}{n} x^n$$ and the limit is just $\binom{\frac{1}{2}}{3}=\frac 1 {16}$ (we do not need to care about the $x^5$ term in denominator since $x^5<<x^3$ when $x\to 0$).

Answer 2

HINT:

$$f(x)=(1+x)^{1/2}=1+\frac12 x-\frac18x^2+\frac1{16}x^3+O(x^4)$$