Problem 17 - Thomas Calculus

Question

Show that \(\sqrt{10 x+1}\) and \(\sqrt{x+1}\) grow at the same rate as \(x \rightarrow \infty\) by showing that they both grow at the same rate as \(\sqrt{x}\) as \(x \rightarrow \infty\) .

Step-by-Step Solution

Answer

Both expressions grow at the same rate as \(x \rightarrow \infty\) with respect to \(\sqrt{x}\).

Step 1: Define Growth to Infinity

To show that the expressions grow at the same rate as \(x \rightarrow \infty\), compare their limits with \(\sqrt{x}\). We need to evaluate: \( \lim_{{x \to \infty}} \frac{\sqrt{10x+1}}{\sqrt{x}} \) and \( \lim_{{x \to \infty}} \frac{\sqrt{x+1}}{\sqrt{x}} \) to see if they both approach a constant.

Step 2: Evaluate First Limit

Calculate \( \lim_{{x \to \infty}} \frac{\sqrt{10x+1}}{\sqrt{x}} \):\[\frac{\sqrt{10x+1}}{\sqrt{x}} = \sqrt{\frac{10x+1}{x}} = \sqrt{10 + \frac{1}{x}}\]As \(x \to \infty\), \(\frac{1}{x}\) approaches 0, so:\[\lim_{{x \to \infty}} \sqrt{10 + \frac{1}{x}} = \sqrt{10}\]

Step 3: Evaluate Second Limit

Calculate \( \lim_{{x \to \infty}} \frac{\sqrt{x+1}}{\sqrt{x}} \):\[\frac{\sqrt{x+1}}{\sqrt{x}} = \sqrt{\frac{x+1}{x}} = \sqrt{1 + \frac{1}{x}}\]As \(x \to \infty\), \(\frac{1}{x}\) approaches 0, so:\[\lim_{{x \to \infty}} \sqrt{1 + \frac{1}{x}} = \sqrt{1} = 1\]

Step 4: Compare Growth Rates with \(\sqrt{x}\)

We have the limits:- \( \lim_{{x \to \infty}} \frac{\sqrt{10x+1}}{\sqrt{x}} = \sqrt{10} \)- \( \lim_{{x \to \infty}} \frac{\sqrt{x+1}}{\sqrt{x}} = 1 \)Since both limits lead to finite constants, \(\sqrt{10x+1}\) and \(\sqrt{x+1}\) grow proportionally to \(\sqrt{x}\) as \(x \rightarrow \infty\). Therefore, they grow at the same rate relative to \(\sqrt{x}\) and hence at the same rate relative to each other.

Key Concepts

Understanding Limits as x Approaches Infinity

When dealing with functions that grow very large as the variable increases, we use the concept of limits to analyze their growth behavior. In mathematics, a limit helps us understand what value a function approaches when the variable within it becomes extremely large or small.
For this exercise, we need to verify how the functions \(\sqrt{10x+1}\) and \(\sqrt{x+1}\) behave as \(x\) becomes infinitely large.
To achieve this, we compare their limits to \(\sqrt{x}\) to check if they converge to constant values.

• We calculate the limit of their ratios with \(\sqrt{x}\).
• By doing this, we can determine if they grow at the same rate.

If the resulting limit of these ratios is a constant, it indicates that as \(x\) goes towards infinity, these functions grow proportionally to \(\sqrt{x}\).

The Concept of Infinity in Mathematics

Infinity is not a number but a concept used to express an unbounded quantity. In calculus, we often deal with limits as a variable approaches infinity, representing limitless growth.
When analyzing the function \(\sqrt{10x+1}\) as \(x\) approaches infinity, we consider the ratio \(\frac{\sqrt{10x+1}}{\sqrt{x}}\). This ratio simplifies to \(\sqrt{10 + \frac{1}{x}}\).
As \(x\) moves towards infinity, \(\frac{1}{x}\) diminishes to zero, simplifying our expression to \(\sqrt{10}\).

• This means the growth rate of \(\sqrt{10x+1}\) relative to \(\sqrt{x}\) is constant and equals \(\sqrt{10}\).

Similarly, for \(\sqrt{x+1}\), the ratio \(\frac{\sqrt{x+1}}{\sqrt{x}}\) becomes \(\sqrt{1 + \frac{1}{x}}\). As \(x\) increases indefinitely, this also approaches 1.
Both results show that each function grows proportionally to \(\sqrt{x}\) as \(x\) tends towards infinity.

Exploring the Role of Square Roots

Square roots often appear in growth rate problems because they help relate linear and non-linear growth. Here, we compare how \(\sqrt{10x+1}\) and \(\sqrt{x+1}\) grow in relation to \(\sqrt{x}\) as x increases.For \(\sqrt{10x+1}\), when divided by \(\sqrt{x}\), our expression simplifies to \(\sqrt{10 + \frac{1}{x}}\). This shows that it initially grows faster than \(\sqrt{x}\), due to the multiplicative factor of 10.
However, as \(x\) approaches infinity, this factor stabilizes, yielding a consistent growth rate relative to \(\sqrt{x}\).

• \(\sqrt{10x+1}\) thus has a growth rate that remains proportional as seen in the limit \(\sqrt{10}\).

The function \(\sqrt{x+1}\), by comparison, stabilizes almost immediately to a growth rate of 1 when divided by \(\sqrt{x}\).
This characterizes \(\sqrt{x+1}\) as growing more slowly compared to \(\sqrt{10x+1}\). However, both grow at a consistent rate when measured against \(\sqrt{x}\) over sufficient range, hence confirming their equal growth rate in terms of calculus.