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Chapter 4: Problem 32

Evaluate the following limits. $$\lim _{x \rightarrow 1} \frac{x^{n}-1}{x-1}, n \text { is a positive integer }$$

Short Answer

Expert verified
Function: \(\lim _{x \rightarrow 1} \frac{x^n - 1}{x - 1}\) Answer: The limit of the given function as x approaches 1 is \(n\).

Step by step solution

01

Identify the indeterminate form

Plug in the value of x=1 into the given function: $$ \lim _{x \rightarrow 1} \frac{x^n - 1}{x - 1} = \frac{1^n - 1}{1 - 1}=\frac{0}{0} $$ Since we have an indeterminate form (0/0), we have to factor the expression in the numerator to cancel out \((x - 1)\) in the denominator.
02

Factor the numerator using difference of powers identity

The numerator is the difference of two powers (x^n and 1), and we can factor it as follows: $$ x^n - 1 = (x - 1)(x^{n - 1} + x^{n - 2} + \cdots + x^2 + x^1 + 1) $$ The factored form of the numerator is a well-known identity, which can be derived from the geometric series formula.
03

Cancel out the common factor from the numerator and denominator

Now, we have: $$ \lim _{x \rightarrow 1} \frac{(x - 1)(x^{n - 1} + x^{n - 2} + \cdots + x^2 + x^1 + 1)}{x - 1} $$ We can cancel the common factor \((x - 1)\) from both the numerator and the denominator: $$ \lim _{x \rightarrow 1} (x^{n - 1} + x^{n - 2} + \cdots + x^2 + x^1 + 1) $$
04

Plug in x=1 to evaluate the limit and simplify

Now we substitute the value of x=1 into the simplified expression: $$ \lim _{x \rightarrow 1} (1^{n - 1} + 1^{n - 2} + \cdots + 1^2 + 1^1 + 1) = (1 + 1 + \cdots + 1 + 1 + 1) $$ Since we have n terms in the sum, each equal to 1, the limit evaluates to the sum of all the terms: $$ \lim _{x \rightarrow 1} \frac{x^n - 1}{x - 1} = \boxed{n} $$

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