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Chapter 11: Problem 41

The value of \(\lim _{x \rightarrow \infty} \frac{3^{x+1}-5^{x+1}}{3^{x}-5^{x}}\) is (A) 5 (B) \(\frac{1}{5}\) (C) \(-5\) (D) None of these

Short Answer

Expert verified
(A) 5

Step by step solution

01

Simplify the Expression

Consider the expression \(\frac{3^{x+1} - 5^{x+1}}{3^x - 5^x}\). The terms \(3^{x+1}\) and \(5^{x+1}\) can be rewritten using the property of exponents. Hence, \(3^{x+1} = 3 \cdot 3^x\) and \(5^{x+1} = 5 \cdot 5^x\). Substituting these back into the expression, we have \(\frac{3 \cdot 3^x - 5 \cdot 5^x}{3^x - 5^x}\).
02

Factor Out the Dominant Term

As \(x\) approaches infinity, the term \(5^x\) dominates because it grows faster than \(3^x\). Therefore, we should factor \(5^x\) out from both the numerator and the denominator. The expression becomes \(\frac{5^x(3 \cdot \left(\frac{3}{5}\right)^x - 5)}{5^x(\left(\frac{3}{5}\right)^x - 1)}\). Cancel \(5^x\) from the numerator and the denominator.
03

Evaluate the New Expression as \(x\) Approaches Infinity

The expression simplifies to \(\frac{3 \left(\frac{3}{5}\right)^x - 5}{\left(\frac{3}{5}\right)^x - 1}\). As \(x\) approaches infinity, \(\left(\frac{3}{5}\right)^x\) approaches 0 because \(\frac{3}{5} < 1\). Substitute 0 for \(\left(\frac{3}{5}\right)^x\), resulting in \(\frac{3 \cdot 0 - 5}{0 - 1}\).
04

Simplify the Result

The expression simplifies to \(\frac{0 - 5}{0 - 1} = \frac{-5}{-1} = 5\). Thus, the value of the limit is 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dominant Term
In calculus, especially when dealing with limits as a variable approaches infinity, one term in an expression can often have a significantly larger impact than others. This is what we refer to as the "dominant term."
  • For example, if you have an expression that involves powers of a number, the term with the largest base raised to a power often dominates the others as the variable grows.
  • In the given problem, we compare two terms:
    • \(3^x\) and \(5^x\).
    As \(x\) approaches infinity, \(5^x\) becomes significantly larger than \(3^x\). Therefore, \(5^x\) is the dominant term.
Recognizing the dominant term helps to simplify the expression by either factoring it out or ignoring smaller terms in the comparison. In this problem, factoring out \(5^x\) greatly simplifies the expression, displaying the power of this technique in finding limits.
Exponential Growth
Exponential growth describes situations where quantities increase rapidly based on their size. In mathematical terms, an exponential function is characterized by a constant ratio of growth against another term or function.
  • The term \(3^x\) represents exponential growth, but with a base of 3.
  • The term \(5^x\) also exhibits exponential growth but with a base of 5.
When comparing these exponential functions, the growth rate of the base 5 term outpaces that of the base 3 term.This is because:
  • A higher base in an exponential function means faster growth as the exponent increases.
In practice, realizing which function grows faster allows you to make key simplifications,such as dismissing smaller growth when considering limits towards infinity.
Infinity in Calculus
Infinity in calculus is a concept that deals with what happens to functions and expressions as variables grow larger and larger, beyond any finite limits.
  • When you evaluate a limit as \(x\) approaches infinity, you essentially examine the behavior of the expression for very large values of \(x\).
  • This often involves simplifying expressions to determine what term dictates the overall behavior.
In the problem provided, understanding how \(\left(\frac{3}{5}\right)^x\) behaves as \(x\) becomes very large is crucial:
  • Because \(\frac{3}{5} < 1\), this term approaches zero as \(x\) approaches infinity.
This simplification leads to final results that are easier to handle mathematically, such as determining that after substituting this term into our final expression, we are left with the limit of 5. Thus, using infinity effectively in calculus, allows students to solve limits clearly and efficiently.

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Most popular questions from this chapter

\(\lim _{n \rightarrow \infty} n^{-n^{2}}\left[(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^{2}}\right) \ldots\left(n+\frac{1}{2^{n-1}}\right)\right]^{n}\) (A) \(e\) (B) \(e^{2}\) (C) \(e^{4}\) (D) None of these

Column-I Column-II (I) \(\lim _{n \rightarrow \infty}\left[\sqrt[3]{n^{2}-n^{3}}+n\right]\) (A) \(\frac{1}{9}\) (II) \(\lim _{x \rightarrow 1} \frac{\sqrt[3]{x^{2}}-2 \sqrt[3]{x}+1}{(x-1)^{2}}\) (B) \(\frac{6}{7}\) (III) \(\lim _{n \rightarrow \infty} \prod_{r=3}^{n}\left(\frac{r^{3}-1}{r^{3}+1}\right)\) (C) \(\frac{1}{3}\) (IV) \(\lim _{n \rightarrow \infty}\left(\cos \frac{x}{n}\right)^{n}\) (D) 1

The value of \(\lim _{x \rightarrow \infty} \frac{x^{5}}{5^{x}}\) is (A) 1 (B) \(-1\) (C) 0 (D) None of these

If \(\lim _{x \rightarrow y} \frac{x^{y}-y^{x}}{x^{x}-y^{y}}=\frac{1-k}{1+k}\), then \(k=\) (A) \(\log y\) (B) \(e^{y}\) (C) \(y\) (D) None of these

\(\lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}}=0,(n\) integer \()\), for (A) no value of \(n\) (B) all values of \(n\) (C) only negative values of \(n\) (D) only positive values of \(n\)
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