Question

If $\mathop {\lim }\limits_{x \to 5} \dfrac{{{x^k} - {5^k}}}{{x - 5}} = 500$ , then the positive integral value of is k
A. 3
B. 4
C. 5
D. 6

Answer 1

We will get indeterminate form after putting the limits hence we will use L-Hospital rule, after using L-Hospital rule and differentiating both the numerator and denominator with respect to x we will get $\mathop {\lim }\limits_{x \to 5} k{x^{k - 1}} = 500$ , now we will try to prime factorize 500 and compare LHS and RHS to get our final answer.

Complete step by step solution:
We have,
$\mathop {\lim }\limits_{x \to 5} \dfrac{{{x^k} - {5^k}}}{{x - 5}}$ … (1)

Now, we should first try to put the limits to solve it,

Putting limits in (1), we get
$\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{{x^k} - {5^k}}}{{x - 5}} = \dfrac{{{5^k} - {5^k}}}{{5 - 5}} = \dfrac{0}{0}$

Clearly, we got an indeterminate form
We know that whenever we get indeterminate forms we can apply L-Hospital rule in $\mathop {\lim }\limits_{x \to 5} \dfrac{{{x^k} - {5^k}}}{{x - 5}}$
According to L-Hospital Rule, a limit which evaluates to indeterminate form (i.e $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ ) can be represented as,
$\Rightarrow \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}$ … (2)

So, numerator and denominator of (1) are
Numerator $= f\left( x \right) = ({x^k} - {5^k})$ … (3)
Denominator $= g\left( x \right) = (x - 5)$ … (4)

Now differentiating (3) and (4) with respect to x, we get
$f^{‘}{x}=\dfrac{d}{dx} \left( x^k - 5^k \right)$
SInce $=5^k$ is a constant with respect to x. Hence
$f^{‘}{x}=\dfrac{d}{dx} \left( x^k \right)$
We know that differentiation of $x^n$ is $nx^{n-1}$. Using the same
$f^{‘}{x}= kx^{k-1}$ ....(5)
Now for the denominator we have
$g^{‘}{x}=x-5$
Differentiating g(x) with respect to x, we get
$g^{‘}{x}=\dfrac{d}{dx} \left( x - 5 \right)$
Since, 5 is a constant and we know that $\dfrac{{dx}}{{dx}}$ is 1, we get
$g^{‘}{x}=1$ … (6)
Now, put (5) and (6) in (2) we get
$\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{{x^k} - {5^k}}}{{x - 5}} = \mathop {\lim }\limits_{x \to 5} \dfrac{{k{x^k}}}{1}$
Now, put value of limit to solve it, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{k{x^k}}}{1} = k \times {5^{k - 1}}$
Also, we are given
$\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{{x^k} - {5^k}}}{{x - 5}} = 500$
We get,
$\Rightarrow k \times {5^{k - 1}} = 500$ … (7)
Now to find the value of k, we can prime factorise 500 and compare LHS with RHS,
$\Rightarrow 500 = 4 \times 5 \times 5 \times 5$
Hence, we get
$\Rightarrow 500 = 4 \times {5^3}$ … (8)
Comparing (7) with (8), we get
$\Rightarrow k \times {5^{k - 1}} = 4 \times {5^3}$
We know that $4 = 3 - 1$ , Hence
$\Rightarrow k \times {5^{k - 1}} = 4 \times {5^{4 - 1}}$ … (9)
Clearly, from (9), we get
$\Rightarrow k = 4$

Hence, the correct option is B.

Note:
In this question, differentiate with respect to x because we can easily get confused here and may differentiate with respect to k.
This Question has a direct formulae application (Alternate way).
If we come across a limit where,
$\mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}}$ ,
Then the direct answer to this limit is
$\mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n \times {x^{n - 1}}$
This can be remembered and it will surely help in other similar questions.