Question

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

Answer 1

Hint: Use the L’ Hospital Rule to solve the limit as the given function is in $\dfrac{0}{0}$ indeterminate form when the value of x is substituted in the function. In this rule we differentiate the numerator and differentiate the denominator then put the limit.

Complete step-by-step answer:
Given function $\mathop {{\text{lim}}}\limits_{{\text{x}} \to {\text{5}}} \dfrac{{{{\text{x}}^{\text{k}}}{\text{ - }}{{\text{5}}^{\text{k}}}}}{{{\text{x - 5}}}} = 500$
Here we have to find the value of k.
When we substitute the value of x in the function we get,
$ \Rightarrow \mathop {{\text{lim}}}\limits_{{\text{x}} \to {\text{5}}} \dfrac{{{{\text{x}}^{\text{k}}}{\text{ - }}{{\text{5}}^{\text{k}}}}}{{{\text{x - 5}}}} = \dfrac{{{5^{\text{k}}}{\text{ - }}{{\text{5}}^{\text{k}}}}}{{{\text{5 - 5}}}} = \dfrac{0}{0}$
The function is in $\dfrac{0}{0}$ indeterminate form as it does not give us the value of limit.
So we will use the L' Hospital rule. Here, we differentiate the numerator and differentiate the denominator separately, then put the limit. This means that the limit of indeterminate form is equal to the limit of derivative of the function.
We know that $\dfrac{{d\left( {{{\text{x}}^{\text{n}}}} \right)}}{{dx}} = {\text{n}}{{\text{x}}^{{\text{n - 1}}}}$ . So on using this formula we get-
$ \Rightarrow \mathop {{\text{lim}}}\limits_{{\text{x}} \to {\text{5}}} \dfrac{{{{\text{x}}^{\text{k}}}{\text{ - }}{{\text{5}}^{\text{k}}}}}{{{\text{x - 5}}}} = \mathop {{\text{lim}}}\limits_{{\text{x}} \to {\text{5}}} \dfrac{{{\text{k}}{{\text{x}}^{{\text{k - 1}}}}{\text{ - 0}}}}{{\text{1}}} = 500$ as differentiation of constant is zero.
$ \Rightarrow \mathop {{\text{lim}}}\limits_{{\text{x}} \to {\text{5}}} \dfrac{{{\text{k}}{{\text{x}}^{{\text{k - 1}}}}}}{1} = 500$
Now, we will put the value of x in the equation.
$ \Rightarrow {\text{k}}{\left( 5 \right)^{{\text{k - 1}}}} = 500$
Now we have to break $500$ into its factors to simplify the equation.
So we can write $500 = 4 \times 125$
On substituting this value in the equation we get-
$ \Rightarrow {\text{k}}{\left( 5 \right)^{{\text{k - 1}}}} = 4 \times 125$
Now we can write $125 = 5 \times 5 \times 5 = {5^3}$ and we have to put the value in form of ${{\text{x}}^{{\text{k - 1}}}}$ so that we can get the value of k. We can write ${5^3} = {5^{4 - 1}}$ then substituting this value in the equation we get,
$ \Rightarrow {\text{k}}{\left( 5 \right)^{{\text{k - 1}}}} = 4 \times {5^{4 - 1}}$
From the above equation we can see that value of k=$5$
Hence, the correct answer is ‘C’.

Note: L’ Hospital rule is used to solve the limits of indeterminate forms. The indeterminate forms are of the following types-$\dfrac{0}{0},\dfrac{\infty }{\infty },\dfrac{0}{{ - \infty }},{1^\infty },{0^0},{\infty ^0},\infty - \infty $ as we cannot determine its exact value. L’ Hospital rule converts the indeterminate form to such expression which can be easily evaluated by substituting the value of given limit.