Question

Find the value of $k$, if $\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$

Answer 1

Here, we are required to find the value of $k$, when we are given two limits which are equal to each other. We will solve both the limits separately using some mathematical formulas in the numerators and denominators. We will then substitute the limit in the place of $x$ to get the value of both the LHS and RHS. Then we will equate these values and solve it further to get the required value of $k$.

Formula Used:
We will use the following formulas:

1. ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
2. $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$

Complete Step by step Solution:
First, we will find the value of LHS i.e. $\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}}$.
Also, we can write this as:
$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{{{\left( {{x^2}} \right)}^2} - {1^2}}}{{x - 1}}$
Now, using the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the numerator, we get,
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}{{x - 1}}$
Again, using the same formula, ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ again in the numerator, we get,
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)}}{{x - 1}}$
Cancelling out $\left( {x - 1} \right)$ from the numerator and denominator, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right)\left( {{x^2} + 1} \right)$
Now, substituting $x = 1$ in the above equation, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \left( {1 + 1} \right)\left( {1 + 1} \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = 2 \times 2 = 4$
Therefore, LHS $= \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = 4$
Now, we will solve the RHS, i.e. $\mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$.
Here, using the formula $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$ in the numerator, we get
$\Rightarrow \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \mathop {\lim }\limits_{x \to k} \dfrac{{\left( {x - k} \right)\left( {{x^2} + xk + {k^2}} \right)}}{{{x^2} - {k^2}}}$
Also, using the formula ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ in the denominator, we get, $\Rightarrow \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \mathop {\lim }\limits_{x \to k} \dfrac{{\left( {x - k} \right)\left( {{x^2} + xk + {k^2}} \right)}}{{\left( {x - k} \right)\left( {x + k} \right)}}$
Cancelling out the same terms from the numerator and denominator, we get
$\Rightarrow \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \mathop {\lim }\limits_{x \to k} \dfrac{{\left( {{x^2} + xk + {k^2}} \right)}}{{\left( {x + k} \right)}}$
Now, substituting $x = k$ in the above equation, we get
$\Rightarrow \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \dfrac{{\left( {{k^2} + \left( k \right)\left( k \right) + {k^2}} \right)}}{{\left( {k + k} \right)}}$
$\Rightarrow \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}} = \dfrac{{3{k^2}}}{{2k}}$
Now, according to the question, $\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^4} - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to k} \dfrac{{{x^3} - {k^3}}}{{{x^2} - {k^2}}}$.
Therefore, equating their values, we get
$4 = \dfrac{{3{k^2}}}{{2k}}$………………………..$\left( 1 \right)$
On cross multiplication, we get
$\Rightarrow 8k = 3{k^2}$
$\Rightarrow 3{k^2} - 8k = 0$
Taking $k$ common, we get
$\Rightarrow k\left( {3k - 8} \right) = 0$
Using zero product property, we get
$k = 0$
Or
$3k - 8 = 0$
Adding 8 on both sides, we get
$\Rightarrow 3k = 8$
Dividing both sides by 3, we get
$\Rightarrow k = \dfrac{8}{3}$
Now, substituting $k = 0$ in equation $\left( 1 \right)$, we get
$4 \ne \dfrac{{3\left( 0 \right)}}{{2\left( 0 \right)}} \ne 0$
We can see that LHS $\ne$ RHS.
Hence, the value of $k = 0$ is rejected.
Now substituting $k = \dfrac{8}{3}$ in equation $\left( 1 \right)$, we get
$\Rightarrow 4 = \dfrac{{3{{\left( {\dfrac{8}{3}} \right)}^2}}}{{2\left( {\dfrac{8}{3}} \right)}}$
Applying exponent on the terms, we get
$\Rightarrow 4 = \dfrac{{3\left( {\dfrac{{64}}{9}} \right)}}{{2\left( {\dfrac{8}{3}} \right)}}$
Simplifying the expression, we get
$\Rightarrow 4 = \dfrac{{\left( {\dfrac{{64}}{3}} \right)}}{{\left( {\dfrac{{16}}{3}} \right)}} = \dfrac{{64}}{{16}} = 4$
Here, LHS $=$ RHS

Hence, the value of $k = \dfrac{8}{3}$.
Therefore, this is the required answer.

Note:
In mathematics, a limit is a value that a function approaches as the input or the index approaches some value. Limits are an essential element of calculus and are used to define continuity, integrals and derivatives. ‘Archimedes of Syracuse’ first developed the idea of limits in the third century B.C. to measure curved figures and the volume of a sphere. Limits are always applied at the end and not at the beginning because if it is applied in the beginning then we will get either 0 or infinity, which will be an incorrect answer.