
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
484.8k+ views
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.
* 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.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
