Scary Limit

differentiate the numerator and denominator .....at last for the limit to exist and to differentiate one more we will have to take $\alpha$ as 2... at the end we obtain a constant as x is eliminated in the expression of limit.. we get 0.01041 add 2 to it

$\text{We use binomial expansion to simplify the given expression:}$

$\large\varphi= \displaystyle\lim_{x\to 0}\dfrac{\alpha - \alpha\sqrt{1-\frac{x^{2}}{\alpha^{2}}}-\frac{x^{2}}{4}}{x^{4}} = \displaystyle\lim_{x\to 0}\dfrac{\alpha - \alpha + \alpha\frac{1}{2}\frac{x^{2}}{\alpha^{2}} - \alpha\frac{1}{2}(\frac{1}{2}-1)\frac{x^{4}}{\alpha^{4}}+\Psi(x) - \frac{x^{2}}{4}}{x^{4}}$

$\text{Where }\Psi(x) \text{ is a polynomial in x with all the exponents greater than 4.}$

$\text{Thus, when we take out }\Psi(x)\text{ by distributive law of division, we will get a function in x}$

$\text{(an infinitesimal of order greater than 2), which will ultimately amount to 0 as the limit is applied.}$

$\therefore\text{we basically have to evaluate:}$

$\large\displaystyle\lim_{x\to 0}\dfrac{\frac{1}{2}\frac{x^{2}}{\alpha} + \frac{1}{4}\frac{x^{4}}{\alpha^{3}} - \frac{x^{2}}{4}}{x^{4}}$

Now, since $\varphi$ is finite, the coefficient of $x^{2}$ in the numerator should be $=0$

Thus, by solving, we get: $\alpha=2$ and the value of the limit:

$\large\varphi=\dfrac{1}{4\alpha^{3}}=\boxed{\dfrac{1}{32}}$

$\large\therefore\varphi+\alpha=\large\boxed{2.03125}$