First slide

Introduction to limits

Question

$lim_{x \to \infty} \frac{x^{2^{32}} - 2^{32} x + 4^{16} - 1}{(x - 1)^{2}}$ is equal to

Moderate

A

$2^{63} - 2^{31}$
B

$2^{64} - 2^{31}$
C

$2^{62} - 2^{31}$
D

$2^{65} - 2^{33}$

Solution

$lim_{x \to 1} \frac{x^{23} - 2^{32} x + 4^{16} - 1}{(x - 1)^{2}}$
$= lim_{x \to 1} \frac{x^{n} - n x + n - 1}{(x - 1)^{2}},$ where $n = 2^{32}$
$\begin{array}{l} = lim_{x \to 1} \frac{(x^{n} - 1) - n (x - 1)}{(x - 1)^{2}} \\ = lim_{x \to 1} \frac{(x^{n - 1} + x^{n - 2} + x^{n - 3} + \dots + x + 1) - n}{x - 1} \end{array}$
$\begin{array}{l} = lim_{x \to 1} \frac{(x^{n - 1} - 1) + (x^{n - 2} - 1) + (x^{n - 3} - 1) + \dots + (x - 1)}{x - 1} \\ = lim_{x \to 1} (\frac{x^{n - 1} - 1}{x - 1} + \frac{x^{n - 2} - 1}{x - 1} + \frac{x^{n - 3} - 1}{x - 1} + \dots + \frac{x - 1}{x - 1}) \\ = {(n - 1) + (n - 2) + (n - 3) + \dots + 2 + 1} \end{array}$
$= \frac{n (n - 1)}{2} = 2^{31} (2^{32} - 1) = 2^{63} - 2^{31}$

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