Question

Evaluate: \[\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{15}} - 1}}{{{x^{10}} - 1}}\].

Answer 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.

Complete step by step answer:

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.