Problem 231 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow \infty} \frac{(x+1)^{10}+(x+2)^{10}+\ldots+(x+100)^{10}}{x^{10}+10^{10}}\\{\text { Ans. } 100\\} $$

Step-by-Step Solution

Answer

The simplified limit expression becomes \(\lim_{x \to \infty} \frac{100+\frac{\text{(terms with lower powers of } x)}{x^{10}}}{1+\frac{10^{10}}{x^{10}}}\). As \(x\) approaches infinity, the terms with lower powers of \(x\) and the term \(\frac{10^{10}}{x^{10}}\) tend to 0. Therefore, the final answer is \(\boxed{100}\).

Step 1: Identify the sum

Notice that the function has the sum of powers of \(x\) in the numerator: \((x+1)^{10}+(x+2)^{10}+\cdots+(x+100)^{10}\) #Step 2: Expand the sum in the numerator and combine like terms#

Step 2: Expand the sum

Using Binomial Theorem or by simplifying the sum, we get \(100x^{10}+\text{(terms with lower powers of } x)\) #Step 3: Simplify the expression and write it in terms of powers of x#

Step 3: Simplify the expression

The expression becomes: \(\frac{100x^{10}+\text{(terms with lower powers of } x)}{x^{10}+10^{10}}\) #Step 4: Determine the highest power of x in both numerator and denominator#

Step 4: Consider the highest power of x

In the simplified expression, the highest power of \(x\) in both the numerator and denominator is \(x^{10}\). #Step 5: Divide the numerator and denominator by x to the highest power (i.e., x^10)#

Step 5: Divide by the highest power of x

Divide both the numerator and the denominator by \(x^{10}\) to get: \(\frac{100+\frac{\text{(terms with lower powers of } x)}{x^{10}}}{1+\frac{10^{10}}{x^{10}}}\) #Step 6: Use limit properties to evaluate the limit as x approaches infinity#

Step 6: Evaluate the limit

The limit expression becomes: \(\lim_{x \to \infty} \frac{100+\frac{\text{(terms with lower powers of } x)}{x^{10}}}{1+\frac{10^{10}}{x^{10}}}\) As \(x\) approaches infinity, the terms with lower powers of \(x\) in the numerator and the term \(\frac{10^{10}}{x^{10}}\) in the denominator would tend to 0. So, the limit expression simplifies to: \(\lim_{x\to\infty} \frac{100}{1}\) #Step 7: Determine the final answer#

Step 7: Final answer

The limit as \(x\) approaches infinity is \(\boxed{100}\).

Key Concepts

Binomial Theorem

Expanding expressions that involve powers, such as \((x+k)^{10}\), often require the use of the Binomial Theorem.
This theorem provides a way to express \((a+b)^n\) as an expanded sum involving terms of the form \(\binom{n}{k}a^{n-k}b^k\).

• In our problem, each term like \((x+1)^{10}\) is expanded using the Binomial Theorem.
• This results in the dominant term being \(x^{10}\), since it is the highest power when expanded for large values of \(x\).

This powerful theorem helps convert complex polynomial expressions into a sum that highlights the contribution of each power. For enormous values of \(x\), terms involving lower powers become less significant.

Higher Powers of x

In the analysis of limits involving polynomials, recognizing the highest power of \(x\) is crucial.
Higher powers of \(x\) dominate the behavior of expressions as \(x\) approaches infinity.

• For example, in the expression \((x+1)^{10} + (x+2)^{10} + \ldots + (x+100)^{10}\), the term \(x^{10}\) is the leading term for each part due to its exponent.
• The sum \(100x^{10}\) in the numerator essentially drowns out any of the smaller powered terms, meaning they become negligible.
• Similarly, in the denominator \(x^{10} + 10^{10}\), \(x^{10}\) will control the growth of the expression.

Thus, when simplifying, it’s often necessary to identify and focus on the behavior of these dominant terms and recognize that lower-powered terms will diminish in influence as \(x\) gets very large.

Asymptotic Behavior

Understanding an expression's asymptotic behavior is key to evaluating limits like this one as \(x\) tends to infinity.

When we divide both the numerator and the denominator by the highest power of \(x\) in the expression, here \(x^{10}\), we expose the ratio of the leading coefficients:

• The numerator \(\frac{100x^{10}}{x^{10}}\) leads to the simplified term \(100\).
• The denominator becomes \(\frac{x^{10} + 10^{10}}{x^{10}}\), simplifying to \(1 + \frac{10^{10}}{x^{10}}\).

As \(x\) increases indefinitely, \(10^{10}/x^{10}\) decreases towards zero, leaving us with the core behavior of \(\frac{100}{1}\).

Thus, this asymptotic analysis reveals that the limit is \(100\), demonstrating the importance of focusing on the primary behavior as smaller terms fade away.