if   \lim_{x\rightarrow 2}\frac{tan(x-2)\left \{ x^{2}+(k-2)x-2k \right \}}{x^{2}-4x+4}=5,   then k is equal to :

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Answers (1)

As we learnt in

Evalution of Trigonometric limit -

\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1

\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1

put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0

Then\:it\:comes

\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1

\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and

\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1

-

 

 \lim_{x\rightarrow 2}\:\:\: \frac{\tan (x-2)\:(x^{2}+(k-2)\:x-2k)}{x^{2}-4x+4}=5

\lim_{x\rightarrow 2}\:\:\: \frac{\tan (x-2)\:(x^{2}+kx-2x-2k)}{(x-2)^{2}}

\lim_{x\rightarrow 2}\:\:\: \frac{\tan (x-2)}{(x-2)}\:\:.\:\:\frac{(x+k)(x-2)}{(x-2)}

x+k=5

k+2=5

\therefore k=3


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