Evaluate $\lim_\limits{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1}$ For all $n,k \in N a,b > 0$
$$\lim_{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1} = \lim_{x \to 0}\frac{x}{(1+ \frac{ax}{n})(1+ \frac{bx}{k})- 1}= \lim_{x \to 0}\frac{x}{x(\frac{a}{n} + \frac{b}{k}) + \frac{ab}{nk}x^2} = \lim_{x \to 0}\frac{1}{\frac{a}{n} + \frac{b}{k}} = \frac{nk}{ka+nb}$$ 
Am I right?

The limit is fine. Perhaps with a view to formalizing the procedure is appropriate to introduce an infinite series , and then Landau notation.
$$O(g(x)) = \left\{\begin{matrix} f(x) : \forall x\ge x_0 >0 , 0\le |f(x)|\le c|g(x)| \end{matrix}\right\}$$
We will also use:
$$f_1=O(g_1)\wedge f_2=O(g_2)\implies f_1f_2=O(g_1g_2)\,$$
Recalling, from MacLaurin series:
$${(1+ax)^{1/n}}={\sum_{i=0}^{\infty} \binom{1/n}{i} (ax)^i }={ (1+\dfrac{ax}{n}+O(x^2))}\\~\\
{(1+bx)^{1/k}}={( 1+\dfrac{bx}{k}+O(x^2))}
$$
Now, we have:
$${(1+ax)^{1/n}}{(1+bx)^{1/k}}={ (1+\dfrac{ax}{n}+O(x^2))}{( 1+\dfrac{bx}{k}+O(x^2))}={(1+\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}
$$
And finally
$$
\lim_{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1} = \lim_{x \to 0}\frac{x}{{(1+\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}-1}\\~\\=  \lim_{x \to 0}\frac{(x^{-1})x}{{(x^{-1})(\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}} = \lim_{x \to 0}\frac{1}{\frac{a}{n} + \frac{b}{k} +O(x)} = \frac{1}{\frac{a}{n} + \frac{b}{k}}= \frac{nk}{ak+nb}
$$
