Q97 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { The value of } \lim _\limits{x \rightarrow 1} \dfrac{x^{15}-1}{x^{10}-1}=$
- A1
- BDoes not exist
- C$\dfrac{2}{3}$
- Correct answerD$\dfrac{3}{2}$
Solution
The limit is of the form $\dfrac{0}{0}$ as $x \rightarrow 1$.
Using L'Hopital's rule, we differentiate the numerator and the denominator with respect to $x$:
$\lim_{x \rightarrow 1} \dfrac{\dfrac{d}{dx}(x^{15} - 1)}{\dfrac{d}{dx}(x^{10} - 1)} = \lim_{x \rightarrow 1} \dfrac{15x^{14}}{10x^9}$
Simplifying the expression:
$\lim_{x \rightarrow 1} \dfrac{15}{10} x^{14-9} = \dfrac{3}{2} \lim_{x \rightarrow 1} x^5$
Substituting $x = 1$:
$\dfrac{3}{2} \times (1)^5 = \dfrac{3}{2}$
Answer: $\dfrac{3}{2}$
Using L'Hopital's rule, we differentiate the numerator and the denominator with respect to $x$:
$\lim_{x \rightarrow 1} \dfrac{\dfrac{d}{dx}(x^{15} - 1)}{\dfrac{d}{dx}(x^{10} - 1)} = \lim_{x \rightarrow 1} \dfrac{15x^{14}}{10x^9}$
Simplifying the expression:
$\lim_{x \rightarrow 1} \dfrac{15}{10} x^{14-9} = \dfrac{3}{2} \lim_{x \rightarrow 1} x^5$
Substituting $x = 1$:
$\dfrac{3}{2} \times (1)^5 = \dfrac{3}{2}$
Answer: $\dfrac{3}{2}$
Hence, the correct answer is option: 'D'.