Question

lim x→2 tan(x-2)(x2+(k-2)x-2k) = 5

x2-4x+4

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$k=3$

Step-by-step explanation:

$\lim_{x\rightarow 2}\frac{(tan(x-2))\cdot (x^2+(k-2)x-2k}{x^2-4x+4}=5$

When we put the value of x=2 then

$\frac{0}{0}$ form

Apply L'hospital rule

$\lim_{x\rightarrow}\frac{sec^2(x-2)(x^2+(k-2)x-2k)+(2x+k-2)tan(x-2)}{2x-4}$

Again it form $\frac{0}{0}$

Again L'hospital rule

$\lim_{x\rightarrow 2}\frac{2sec(x-2)tan(x-2)(x^2+(k-2)x-2k)+sec^2(x-2)(2x+k-2)+2tan(x-2)+(2x+k-2)sec^2(x-2)}{2}$

$\frac{4+k-2+(4+k-2)}{2}=5$

$\frac{8-4+2k}{2}=5$

$4+2k=10$

$2k=10-4=6$

$k=\frac{6}{2}=3$

$k=3$

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