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d1i1m1o1n [39]
3 years ago
5

How do you show the limit does not exist limx→4x−4x2−8x+16lim_(x->4)(x-4)/(x^2-8x+16) ?

Mathematics
1 answer:
fredd [130]3 years ago
5 0
\lim\limits_{x\to4}\frac{x-4}{x^2-8x+16}=\lim\limits_{x\to4}\frac{x-4}{(x-4)^2}=\lim\limits_{x\to4}\frac{1}{x-4}\\\\\lim\limits_{x\to4^-}\frac{1}{x-4}=\frac{1}{0^-}=-\infty\\\\\lim\limits_{x\to4^+}\frac{1}{x-4}=\frac{1}{0^+}=\infty
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aksik [14]

Check the picture below.

we know that AL is an angle bisector, so the angle at A gets cuts into two equal halves, we also know the angle at B is 30°, so the triangle ABC is really a 30-60-90 triangle, meaning the angle at A is really a 60° angle, cut in two halves gives us 30° and 30° as you see in the picture.

if the angles at A and B, inside the triangle ABL, are equal, twin angles are only made in an isosceles by twin sides, that means that AL = BL.

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nikdorinn [45]

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When we know all 3 sides, we use Heron's Formula

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Step-by-step explanation:

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3 years ago
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vlabodo [156]

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2 years ago
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dolphi86 [110]

Answer:

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5 0
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