<s>[INST] $\lim_{x\to 0^+}\dfrac{\tan^2(\frac1{\sqrt{1-x^2}}-1)}{\left(1-\cos(\sqrt{2x})\right)^n}=a$. What is the value of $a+n$? 
Suppose $\lim_{x\to
 0^+}\dfrac{\tan^2(\frac1{\sqrt{1-x^2}}-1)}{\left(1-\cos(\sqrt{2x})\right)^n}=a$.
What is the value of $a+n$ ?
$1)\frac74\qquad\qquad2)\frac94\qquad\qquad3)\frac{15}4\qquad\qquad4)\frac{17}4$

The limit is $\frac00$. For $x\to0^+$ both numerator and denominator tends to $0^+$, so $a>0$. I can use equivalences for $\tan u\sim u$ For  $u\to0$. Hence in the numerator we can write $\dfrac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}}$. But I'm not sure if this helps.
I think I should find $n$ first then evaluate the limit but I don't know how to find $n$ at first place.
 [/INST] We have, from Newton's series:
$$(1-x^2)^{-\frac 12}=1+\frac {x^2}{2}+O(x^4)$$
Also, Taylor series for $\tan x$ yields:
$$\tan x=x+\frac {x^3}{3}+O(x^5)$$
Hence, $$\tan\left(\frac {1}{\sqrt {1-x^2}}-1\right)=\frac {x^2}{2}+O(x^4)$$
Squaring results in:
$$\tan^2\left(\frac {1}{\sqrt{1-x^2}}-1\right)=\frac {x^4}{4}+O(x^6)$$
Now, Taylor Series for $$\cos(\sqrt {2x})=1-\frac {(\sqrt {2x})^2}{2!}+\frac {(\sqrt {2x})^4}{4!}+O(x^3)$$
Thus, denominator evaluates to:
$$(1-\cos(\sqrt{2x}))^n=(x-\frac {x^2}{6}+O(x^3))^n$$
Thus, it is clear that non-zero limiting value
$$L=\frac {\frac {x^4}{4}+O(x^6)}{x^n(1-\frac {x}{6}+O(x^2))^n}$$
exists at $n=4$, and at that value $a=\frac 14$. Hence, we get option D.
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