Questions & Answers
Question
Answers

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. \[0\]

Answer Verified Verified
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. \[\mathop {\lim }\limits_{x \to a} f(x) = f(a)\]
* \[\mathop {\lim }\limits_{x \to p} (xy) = \mathop {\lim }\limits_{x \to p} x \times \mathop {\lim }\limits_{x \to p} y\]
* \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1\]

Complete step-by-step answer:
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.

Bookmark added to your notes.
View Notes
×