Continuity Problem

Applying the three conditions to make a function continuous at a value a,

$1^{st}$ condition: $\large \lim_{x \rightarrow a} f(x)$ exists.

$\large \begin{aligned} f(x) = \cfrac{x^{2^{32}} - (4^{16})x + 4^{16} - 1}{(x - 1)^2} \ , \ a = 1 \\ \implies \lim_{x \rightarrow 1} \cfrac{x^{2^{32}} - (4^{16})x + 4^{16} - 1}{(x - 1)^2} & = \cfrac{\lim_{x \rightarrow 1} \left( x^{2^{32}} - (4^{16})x + 4^{16} - 1 \right) }{\lim_{x \rightarrow 1} (x - 1)^2} \\ & = \color{#D61F06} {\cfrac{\lim_{x \rightarrow 1} \left( \left(2^{32}\right) x^{2^{32} - 1} - 4^{16} \right) }{\lim_{x \rightarrow 1} \Big( 2(x-1) \Big) }} \\ & = \color{#3D99F6} {\cfrac{\lim_{x \rightarrow 1} \left( \left(2^{32}\right) \left(2^{32} - 1\right) x^{2^{32} - 2} \right)}{\lim_{x \rightarrow 1} (2)}} \\ & = \lim_{x \rightarrow 1} \Big( \left(2^{31}\right) \left(2^{32} - 1\right) x^{2^{32} - 2} \Big) \\ & = \left(2^{31}\right) \left(2^{32} - 1\right) \end{aligned}$

Note: $\large f(x) = \cfrac{x^{2^{32}} - (4^{16})x + 4^{16} - 1}{(x - 1)^2}$ was used as we are dealing with limits.

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$2^{nd}$ condition: $f(a)$ exist:

$f(x) = k \ , \ a = 1 \\ \implies f(1) = k$

Note: $f(x) = k$ was used as we are dealing with the function value.

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$3^{rd}$ condition: $\large \lim_{x \rightarrow a} f(x) = f(a)$

$\large \lim_{x \rightarrow 1} f(x) = \left(2^{31}\right) \left(2^{32} - 1\right) \ , \ f(1) = k \\ \implies k = \left(2^{31}\right) \left(2^{32} - 1\right) \\ \implies a = 2 \ , \ b = 31 \ , \ c = 32 \\ \therefore \ \boxed{a + b + c = 2 + 31 + 32 = 65}$

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The red and blue colored ones was obtain by applying L'Hôpital's Rule .