Those numbers are quite big

Let $\alpha = 432394829$ , $\beta = 3427954283$ , $f(x) = \ln {\left( \tan{ \left( \frac {\pi} {4} + \alpha x\right)}\right)}$ and $g(x) = \sin{\beta x}$ . We note that $\displaystyle \lim_{x \to 0} {f(x)} = \lim_{x \to 0} {g(x)} = 0$ . Therefore, we can use L'Hôspital's rule. $\displaystyle \lim_{x \to 0} {\frac {f(x)}{g(x)}} = \lim_{x \to 0} {\frac {f'(x)}{g'(x)}}$ .

From:

$\begin{aligned} \tan{ \left( \frac {\pi} {4} + \alpha x\right)} & = \dfrac {1+\tan{\alpha x}} {1-\tan{\alpha x}} = \dfrac {1+\tan{\alpha x}} {1-\tan{\alpha x}} \times \dfrac {1+\tan{\alpha x}} {1+\tan{\alpha x}} \\ & = \dfrac {1+2\tan{\alpha x}+\tan^2{\alpha x}} {1-\tan^2{\alpha x}} = \tan{2\alpha x} + \sec{2\alpha x} \end{aligned}$

Therefore,

$\begin{aligned} f(x) & = \ln{(\tan{2\alpha x} + \sec{2\alpha x})} \\ \Rightarrow f'(x) & = \frac {2\alpha\sec^2{2\alpha x}+2\alpha \tan{2\alpha x} \sec{2\alpha x}} {\tan{2\alpha x} + \sec{2\alpha x}} = 2\alpha\sec{2\alpha x} \end{aligned}$

And we have,

$\begin{aligned} \displaystyle \lim_{x \to 0} {\frac {f(x)}{g(x)}} & = \lim_{x \to 0} {\frac {f'(x)}{g'(x)}} = \lim_{x \to 0} {\frac {2\alpha\sec{2\alpha x}}{\beta\cos{\beta x}}} = \frac {2\alpha}{\beta} \end{aligned}$

$\Rightarrow a+b = 2\alpha + \beta = 2\times 432394829 + 3427954283 = \boxed{4292743941}$

Write a solution.

Here is a detailed proof with series expansions, without what we call in french "equivalents", not sure this translates well... and without l'hospital. Here are many calculation details, but this way of reasoning is actually pretty quick (I solved this in my head, except for checking the co primes...)

Let $\alpha = 432394829, \beta = 3427954283$ . $\tan(\pi/4 + \alpha x) = \frac{1+\tan(\pi/4)\tan(\alpha x)}{\tan(\pi/4)-\tan(\alpha x)} = \frac{1+\tan(\alpha x)}{1-\tan(\alpha x)}$ Therefore: $\ln(\tan(\pi/4 + \alpha x)) = \ln(1+\tan(\alpha x)) - \ln(1-\tan(\alpha x))$ .

We then use the series expansion (for $x$ small enough, typically $x<1/(\alpha \beta)$ ):

Moreover, $\sin(\beta x) = \sum_{n=0}^\infty (-1)^n \frac{(\beta x)^{2n+1}}{(2n+1)!} = x + x^3 \sum_{n=1}^\infty (-1)^n \frac{(\beta x)^{2n-1}}{(2n+1)!}$ $\sin(\beta x) = \beta x - (\beta x)^3 B(x)$ $\tan(\alpha x) = \frac{\sum_{n=0}^\infty (-1)^n \frac{(\alpha x)^{2n+1}}{(2n+1)!} }{\sum_{n=0}^\infty (-1)^n \frac{(\alpha x)^{2n}}{(2n)!} } = \frac{\alpha x + \alpha ^3 x^3 \sum_{n=1}^\infty (-1)^n \frac{(\alpha x)^{2n-1}}{(2n+1)!}}{1+\alpha^2 x^2\sum_{n=1}^\infty (-1)^n \frac{(\alpha x)^{2(n-1)}}{(2n)!} }$ $\tan(\alpha x) = \frac{\alpha x - (\alpha x)^3 B(x)}{1-(\alpha x)^2A(x)} = \alpha x(1-(\alpha x)^2 C(x))$ with $C(x) = \frac{B(x)-A(x)}{1-(\alpha x)^2A(x)}$ .

And since $tan(\alpha x) \rightarrow 0$ for $x\rightarrow 0$ , $\ln(1+\tan(\alpha x)) - \ln(1-\tan(\alpha x))=$ $\sum_{n=1}^\infty (-1)^{n+1} \frac{\tan(\alpha x)^{n}}{n} + \sum_{n=1}^\infty \frac{\tan(\alpha x)^{n}}{n} = \sum_{n=1}^\infty 2\frac{\tan(\alpha x)^{2n-1}}{2n-1}$ $= 2 \tan(\alpha x) +\tan(\alpha x)^3 D(\tan(\alpha x))$ $=2 \alpha x + (\alpha x)^3 E(x)$

Of course, $A,B,C,D,E$ are functions that are bounded around $0$ . Putting all this together ... : $\frac{\ln(\tan(\pi/4 + \alpha x))}{\sin(\beta x)} = \frac{2 \alpha x + (\alpha x)^3 E(x) }{\beta x - (\beta x)^3 B(x)} = \frac{2\alpha}{\beta} \frac{1 + (\alpha x)^2 E(x) }{1 - (\beta x)^2 B(x)} \rightarrow \frac{2\alpha}{\beta}$

Checking that $2\alpha$ and $\beta$ are co prime gives the result.