Evaluate $\lim_{x\to 0} {\sqrt[3]{1+{x\over 3}} - \sqrt[4]{1+{x\over 4}} \over 1-\sqrt{1-{x\over 2}}}$

Question

I'm having trouble evaluating this limit:

$$\lim_{x\to 0} {\sqrt[3]{1+{x\over 3}} - \sqrt[4]{1+{x\over 4}} \over 1-\sqrt{1-{x\over 2}}}$$

mainly because I'm not allowed to use L'Hospital's rule, power expansions or derivatives in general.

The main difficulty as I see it is that the arguments of the roots are different, otherwise I could introduce a new variable $f(x) = y^{12}$ and convert to a rational function.

Answer 1

You may prove first that for any $p,q\in\mathbb{N}^+$ $$ \lim_{x\to 0}\frac{\sqrt[p]{1+\frac{x}{q}}-1}{x}=\frac{1}{pq} \tag{1}$$ holds by rationalization, then use such result to show that your limit equals $$ \frac{\frac{1}{9}-\frac{1}{16}}{\frac{1}{4}}=\color{red}{\frac{7}{36}}.\tag{2}$$

Answer 2

We may use binomial expansions to see that

$$\sqrt[3]{1+\frac x3}=1+\frac19x+\mathcal O(x^2)$$

$$\sqrt[4]{1+\frac x4}=1+\frac1{16}x+\mathcal O(x^2)$$

$$\sqrt{1-\frac x2}=1-\frac14x+\mathcal O(x^2)$$

Thus, we have

$$\frac{\sqrt[3]{1+\frac x3}-\sqrt[4]{1+\frac x4}}{1-\sqrt{1-\frac x2}}=\frac{\frac7{144}x+\mathcal O(x^2)}{\frac14x+\mathcal O(x^2)}=\frac{\frac7{144}+\mathcal O(x)}{\frac14+\mathcal O(x)}\to\frac{7\times4}{144}=\boxed{\frac7{36}}$$

Answer 3

Separate thw numerator into two ...-1 terms and work with them individually.

A hint: since $(1+x)^a \approx 1+ax$, the numerator is about x/9-x/16 and the denominator is about 1-(1-x/4).

You can also multiply the num and den by $1+\sqrt{1-x/2}$.