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Question

Answers

A. \[1\]

B. \[2\]

C. \[3\]

D. \[0\]

Answer
Verified

* Limit of a function means the value of the function as it approaches the limit, i.e. \[\mathop {\lim }\limits_{x \to a} f(x) = f(a)\]

* \[\mathop {\lim }\limits_{x \to p} (xy) = \mathop {\lim }\limits_{x \to p} x \times \mathop {\lim }\limits_{x \to p} y\]

* \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]

We have to find the value of k such that \[\mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5\].

First we write the denominator of the fraction in form of its factors. We can write

\[{x^2} - 4x + 4 = {(x)^2} + {(2)^2} - 2(2)(x)\]

This is of the form \[{a^2} + {b^2} - 2ab\]

By comparing the above equation to \[{(a - b)^2} = {a^2} + {b^2} - 2ab\], we can see that the value of \[a = x,b = 2\].

\[{x^2} - 4x + 4 = {(x - 2)^2}\] … (1)

Now we solve the numerator of the fraction.

We have \[\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} \] as the numerator of the fraction.

Opening the brackets by multiplying the terms we get

\[

\Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ {x^2} + kx - 2x - 2k\} \\

\Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ {x^2} - 2x + kx - 2k\} \\

\]

Now take x common from the first two terms and k common from the last two terms in the bracket.

\[ \Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ x(x - 2) + k(x - 2)\} \]

Pairing the factors we get

\[ \Rightarrow \tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} = \tan (x - 2)\{ (x - 2)(x + k)\} \] … (2)

Substituting equation 1 and 2 in given equation, we get

\[ \Rightarrow \dfrac{{\tan (x - 2)\{ (x - 2)(x + k)\} }}{{{{(x - 2)}^2}}} = \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times \dfrac{{(x - 2)(x + k)}}{{(x - 2)}}\]

Cancel out the same terms from numerator and denominator.

\[ \Rightarrow \dfrac{{\tan (x - 2)\{ (x - 2)(x + k)\} }}{{{{(x - 2)}^2}}} = \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times (x + k)\]

Taking limit on both sides we get

\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times (x + k)\]

We know that we can separate the limits as \[\mathop {\lim }\limits_{x \to p} (xy) = \mathop {\lim }\limits_{x \to p} x \times \mathop {\lim }\limits_{x \to p} y\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} \times \mathop {\lim }\limits_{x \to 2} (x + k)\] … (3)

Now we know \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]

So, we can apply this property to the first term because \[x \to 2 = x - 2 \to 0\] and as the function is the same as the limit we can apply the property.

\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)}}{{(x - 2)}} = 1\]

Also, \[\mathop {\lim }\limits_{x \to 2} (x + k) = (k + 2)\]

Substituting the values in equation (3) we get

\[ \Rightarrow \mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 1 \times (k + 2) = k + 2\]

Now from the statement of the question we know the value of LHS is equal to 5.

\[

\Rightarrow k + 2 = 5 \\

\Rightarrow k = 5 - 2 \\

\Rightarrow k = 3 \\

\]

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