Question

Let $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},$ then value of k
( a ) $\dfrac{2}{3}$
( b ) $\dfrac{3}{2}$
( c ) $\dfrac{4}{3}$
( d ) $\dfrac{8}{3}$

Answer 1

To solve this question, what we will do is we will first evaluate the limit of $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}$ using L’ hospital rule and then we will evaluate limit of $\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},$ then we will use relation $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},$ to evaluate value of k.

Complete step-by-step answer:
In question, it is 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, first find the limit of $\dfrac{{{x}^{4}}-1}{x-1}$ when x tends to 1,
So, when $x \to 1$ , $\dfrac{{{x}^{4}}-1}{x-1}$ gives us $\dfrac{0}{0}$ form, which is an indeterminate form.
So to simplify this we will use L’ hopital rule which says that if $\displaystyle \lim_{x \to 1}\dfrac{f(x)}{g(x)}$ gives indeterminate form of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, then we differentiate the functions f ( x ) and g ( x ) with respect to x unless and until we do not get indeterminate form and finally get a finite limit.
So, using L’ hospital rule in $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}$, we get
$\displaystyle \lim_{x \to 1}\dfrac{4{{x}^{3}}}{1}$, as $\dfrac{d}{dx}({{x}^{4}}-1)=4{{x}^{3}}$ and $\dfrac{d}{dx}(x-1)=1$using property of differentiation $\dfrac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$ and $\dfrac{d}{dx}(k)=0$, where k is any constant.
So, $\displaystyle \lim_{x \to 1}4{{x}^{3}}$
Putting x = 1, we get
$4{{(1)}^{3}}=4$……( i )
now, we will evaluate the limit of $\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},$ when x tends to k,
So, $\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},$ it gives us $\dfrac{0}{0}$ form, which is an indeterminate form.
So, using L’ hospital rule in $\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},$, we get
$\displaystyle \lim_{x \to 1}\dfrac{3{{x}^{2}}}{2x}$, as $\dfrac{d}{dx}({{x}^{3}}-{{k}^{3}})=3{{x}^{2}}$ and $\dfrac{d}{dx}({{x}^{2}}-{{k}^{2}})=2x$
On simplifying, we get
$\displaystyle \lim_{x \to 1}\dfrac{3x}{2}$
Putting x = k, we get
$\dfrac{3(k)}{2}=\dfrac{3k}{2}$……( ii )
As in question, it is given that $\displaystyle \lim_{x \to 1}\dfrac{{{x}^{4}}-1}{x-1}=\displaystyle \lim_{x \to k}\dfrac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},$
So from ( i ) and ( ii ), we get
$4=\dfrac{3k}{2}$
Using, cross multiplication, we get
$k=\dfrac{8}{3}$

So, the correct answer is “Option d”.

Note: Questions related algebraic limits, one must know all the rules and short tricks based on evaluation of algebraic limits. Also, while solving indeterminate limits such as $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, L’ hospital rule plays an important role, so on must know the evaluation of limit using L’ hospital rule.