A calculus problem by أحمد الحلاق

Calculus Level 2

$\large \lim_{x\to3} \dfrac{(x^2 - 4x + 3)^{2016}} { (x^2-6x+9)^{1008}} = \, ?$

2^{1008}

2^{5004}

2^{2016}

2^{3032}

1 solution

Chew-Seong Cheong
Sep 7, 2016

$\begin{aligned} L & = \lim_{x \to 3} \frac{(x^2-4x+3)^{2016}}{(x^2-6x+9)^{1008}} \\ & = \lim_{x \to 3} \frac{(x-1)^{2016}(x-3)^{2016}}{\left((x-3)^2 \right)^{1008}} \\ & = \lim_{x \to 3} \frac{(x-1)^{2016}\cancel{(x-3)^{2016}}}{\cancel{(x-3)^{2016}}} \\ & = \boxed{2^{2016}} \end{aligned}$

A nice logical Approach .. +1 :)

Sabhrant Sachan - 4 years, 9 months ago

A calculus problem by أحمد الحلاق

1 solution

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