First slide

Introduction to limits

Question

$The polynomial function f (x) of degree 6 satisfying \underset{x \to 0}{l i m} {\{1 + \frac{f (x)}{x^{3}}\}}^{1 / x} = e^{2} is$

Moderate

A

$x^{4} + 3 x^{5} + 4 x^{6}$
B

$5 x^{4} - 3 x^{5} + 2 x^{6}$
C

$6 x^{4} - 3 x^{5} + x^{6}$
D

$2 x^{4} - 3 x^{5} + 4 x^{6}$

Solution

let $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + a_{5} x^{5} + a_{6} x^{6}$
$\begin{array}{l} \lim_{x \to 0} {\{1 + \frac{f (x)}{x^{3}}\}}^{1 / x} = e^{2} \\ e^{\lim_{x \to 0} \frac{1}{x} (1 + \frac{f (x)}{x^{3}} - 1)} \\ \Rightarrow e^{\lim_{x \to 0} \frac{f (x)}{x^{4}}} = e^{2} \\ \lim_{x \to 0} \frac{f (x)}{x^{4}} = 2 \\ \lim_{x \to 0} \frac{a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + a_{5} x^{5} + a_{6} x^{6}}{x^{4}} = 2 \\ a_{0} = a_{1} = a_{2} = a_{3} = 0, a_{4} = 2, a_{5} =__, a_{6} =__ \end{array}$

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