Question

# If $\mathop {\lim }\limits_{x \to 2} \dfrac{{\tan (x - 2)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5$ then $k$ is equal to A. $1$B. $2$C. $3$D. $0$

Hint: Here we solve the terms in the bracket by converting both numerator and denominator in form of their factors and then cancelling out the same terms to obtain a simpler form. Then we apply the limit to the fraction by breaking the fraction into multiplication of two fractions.
* 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.
$\Rightarrow k + 2 = 5 \\ \Rightarrow k = 5 - 2 \\ \Rightarrow k = 3 \\$

So, the correct answer is “Option C”.

Note: Students are likely to make mistakes while calculating the limit of the tan function and might substitute the value of x as 2 in the fraction which will give us the answer 0 which is wrong. Keep in mind for fractions like $\dfrac{{\tan x}}{x}, \dfrac{{\sin x}}{x}$ we always use this way of finding the limit.