If $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}$, then find the value of k.
(a) $\dfrac{5}{3}$
(b) $-\dfrac{8}{3}$
(c) $\dfrac{8}{3}$
(d) None of these

Hint: We must solve both the limits separately. For this, we must use the expansion formulae ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. By simplification, we can cancel the term that makes the function indeterminate, and then substitute the value of $x$ to calculate the value of k.

Complete step-by-step solution:
We are given that $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}$. Let us solve both of these limits separately.
Let us first solve the limit $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}$.
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Thus, we can write ${{x}^{4}}-1=\left( {{x}^{2}}+1 \right)\left( {{x}^{2}}-1 \right)$.
Again, using the same expansion formula, we can write ${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$.
Thus, we have ${{x}^{4}}-1=\left( {{x}^{2}}+1 \right)\left( x+1 \right)\left( x-1 \right)$.
Hence, the above limit becomes,
$\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\displaystyle \lim_{x \to 1}\dfrac{\left( {{x}^{2}}+1 \right)\left( x+1 \right)\left( x-1 \right)}{x-1}$
And thus, we get
$\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\displaystyle \lim_{x \to 1}\left( {{x}^{2}}+1 \right)\left( x+1 \right)$
Now, we can substitute the value of $x$ as 1 on the right hand side of the above equation. Thus, we have
$\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\left( 1+1 \right)\left( 1+1 \right)$
And so, we get
$\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=4...\left( i \right)$
Let us now solve the limit $\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}$.
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Thus, we can write ${{x}^{2}}-{{k}^{2}}=\left( x+k \right)\left( x-k \right)$.
And we also know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$. Thus, we can write ${{x}^{3}}-{{k}^{3}}=\left( x-k \right)\left( {{x}^{2}}+{{k}^{2}}+xk \right)$
Hence, we can write the above limit as
$\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}=\displaystyle \lim_{x \to k}\dfrac{\left( x-k \right)\left( {{x}^{2}}+{{k}^{2}}+xk \right)}{\left( x+k \right)\left( x-k \right)}$
On cancelling the terms from the numerator and denominator, we get
$\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}=\displaystyle \lim_{x \to k}\dfrac{\left( {{x}^{2}}+{{k}^{2}}+xk \right)}{\left( x+k \right)}$
Now, we can easily substitute the value of $x$ as k on the right hand side of the above equation. Thus, we have
$\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}=\dfrac{\left( {{k}^{2}}+{{k}^{2}}+{{k}^{2}} \right)}{\left( k+k \right)}$
The above equation is equivalent to
$\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}=\dfrac{3{{k}^{2}}}{2k}$
And so, we can now write the above limit as
$\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}}=\dfrac{3k}{2}...\left( ii \right)$
We can see it is given that the left hand side of equations (i) and (ii) are equal to each other. So, the right hand side of these two equations must also be equal.
Thus, we can write
$\dfrac{3k}{2}=4$
Hence, we get
$k=\dfrac{8}{3}$.
Hence, option (c) is the correct answer.

Note: Instead of solving these two limits using the expansion formulae, we can also solve the first limit using the following direct formula, $\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$, and we can solve the second limit using the formula $\displaystyle \lim_{x \to a}\dfrac{{{x}^{m}}-{{a}^{m}}}{{{x}^{n}}-{{a}^{n}}}=\dfrac{m}{n}{{a}^{m-n}}$.