   Question Answers

# Evaluate: $\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{15}} - 1}}{{{x^{10}} - 1}}$.  Hint: First we will first put $x = 1$ in the above equation and check if it is in the $\dfrac{0}{0}$ form. If it is then we will solve the above equation by using the theorem, $\mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$to find the required value.

We are given
$\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{15}} - 1}}{{{x^{10}} - 1}}{\text{ ......eq.(1)}}$

Putting $x = 1$ in the above equation, we get
$\Rightarrow \dfrac{{{1^{15}} - 1}}{{{1^{10}} - 1}} \\ \Rightarrow \dfrac{{1 - 1}}{{1 - 1}} \\ \Rightarrow \dfrac{0}{0} \\$

Since it is of the form $\dfrac{0}{0}$.

Dividing the numerator and denominator of the equation (1) by $x - 1$, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\dfrac{{{x^{15}} - 1}}{{x - 1}}} \right)}}{{\left( {\dfrac{{{x^{10}} - 1}}{{x - 1}}} \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {\dfrac{{{x^{15}} - {1^{15}}}}{{x - 1}}} \right)}}{{\left( {\dfrac{{{x^{10}} - {1^{10}}}}{{x - 1}}} \right)}} \\$
So, we will solve the above equation by using the theorem on limits, $\mathop {\lim }\limits_{x \to a} \dfrac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}$.
$\Rightarrow \dfrac{{15{{\left( 1 \right)}^{14}}}}{{10{{\left( 1 \right)}^9}}} \\ \Rightarrow \dfrac{{15\left( 1 \right)}}{{10\left( 1 \right)}} \\ \Rightarrow \dfrac{{15}}{{10}} \\ \Rightarrow \dfrac{3}{2} \\$

Note: Students should be familiar with the formula to find the limits and the theorems, as some get confused while applying the formulae. When a limit approaches to a number $a$, it does not mean the function is not equal to $a$. We know that any power of 1 is always 1. We will not use the limit on the left equation after removing the variable $x$. Avoid calculation mistakes.
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