Question

\(\operatorname{Lt}_{x \rightarrow \infty} \frac{(x+1)^{100}+(x+2)^{100}+\ldots+(x+50)^{100}}{x^{100}+x^{99}+\ldots+x+1}=\) (1) 100 (2) 1 (3) 21 (4) 50

Short answer

The problem involves finding the limit of a rational function as x approaches infinity. The first step involves dividing both the numerator and denominator by the highest power of x, in this case x^100. Then the expression is simplified, and the limit as x goes to infinity is calculated. Due to the properties of limits, the terms with x in the denominator in the original problem will approach zero, simplifying the expression further. The final step involves counting the number of terms in the numerator and the denominator. The final limit is 50, corresponding to option (4) in the multiple-choice options.

Step-by-step solution

Divide both numerator and denominator by x^100

Divide the given expression by x^100 in the numerator and denominator: $$\lim_{x \rightarrow \infty} \frac{(\frac{x+1}{x})^{100}+(\frac{x+2}{x})^{100}+\ldots+(\frac{x+50}{x})^{100}}{(\frac{x^{100}}{x^{100}})+(\frac{x^{99}}{x^{100}})+\ldots+(\frac{x}{x^{100}})+\frac{1}{x^{100}}}$$.

Simplify the expression

Simplify the terms inside the limit: $$\lim_{x \rightarrow \infty} \frac{(1+\frac{1}{x})^{100}+(1+\frac{2}{x})^{100}+\ldots+(1+\frac{50}{x})^{100}}{(1)+(\frac{1}{x})+(\frac{1}{x^2})+...+\frac{1}{x^{100}}}$$.

Evaluate the limit as x approaches infinity

As x goes to infinity, the terms with x in the denominator will go to zero. After evaluating, we get: $$\lim_{x \rightarrow \infty} \frac{1^{100}+1^{100}+\ldots+1^{100}}{1+0+0\ldots+0}$$.

Count the number of terms

There are 50 terms in the numerator and only one non-zero term in the denominator, so the limit is: $$\frac{50(1)}{1} = 50$$. Hence, the answer is (4) 50.

Key concepts

Indeterminate forms

In calculus, indeterminate forms arise in the evaluation of limits where expressions might not initially appear to converge to a limit definitively. Common indeterminate forms include

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)

When dealing with limits, manipulating these forms into a determinate form allows for clear evaluation.
When addressing the exercise question, the solution initially treats the expression as an indeterminate form since it involves the ratio of two polynomial sum sequences as \( x \rightarrow \infty \).
Mathematicians resolve this by breaking down the polynomial terms to eliminate indeterminate properties.
Thus, by dividing each term by \( x^{100} \), the original indeterminate form becomes approachable.
This manipulation transforms the expressions into forms where the limit can easily be evaluated as infinity approaches.

Asymptotic behavior

Understanding asymptotic behavior involves analyzing how a function behaves as its inputs become very large. In this exercise, as \( x \) approaches infinity, we want to see how the polynomials behave.
The asymptotic analysis helps determine dominant terms in polynomials. For large values of \( x \), terms with the highest power of \( x \) grow faster and dominate the expression.
In the numerator, we look at the sequence from \( (x+1)^{100} \) to \( (x+50)^{100} \). All becoming similar as \( x \) grows indefinitely.
The denominator, \( x^{100} + x^{99} + \ldots + 1 \), simplifies to just \( x^{100} \) as only the highest power term dominates.
This asymptotic understanding simplifies the complex polynomials into manageable forms by focusing on these dominant terms.

Polynomial division

Polynomial division is a critical technique when simplifying expressions in calculus particularly for limits. It involves dividing polynomial expressions to make limits more apparent.
In this problem, polynomial division is used to simplify both the numerator and the denominator.

• By dividing each term by the highest power of \( x \), we break down complex expressions.
• The division simplifies expressions so minute terms become negligible as \( x \rightarrow \infty \).

Through this division, the polynomials transform to linear or constant terms, revealing simple factorial components which are then straightforward to evaluate.
These basic components allow the expression to be taken to the limit effortlessly.

Higher algebra concepts

Higher algebra concepts, including manipulating polynomial expressions and understanding limits, are essential in this calculation.

• They allow calculus problems to be reframed within algebraic terms.
• Facilitate breaking down problems into simpler parts.

For instance, recognizing that every term except the dominant \( x^{100} \) in the denominator approaches zero is key to solving the original expression.
Also, observing each of the 50 binomial numerator terms essentially reduces to \(1\) in the limit as \( x \rightarrow \infty \).
Thus, using these higher algebra techniques ensures even complex limits are handled with precision and ease just like in this exercise.