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Chapter 10: Problem 1

Estimate the limits numerically. \(\lim _{x \rightarrow 0} \frac{x^{2}}{x+1}\)

Short Answer

Expert verified
The limit as x approaches 0 of the function \(\frac{x^2}{x+1}\) is 0, which we can verify analytically and confirm numerically as the function approaches 0 for values very close to 0: \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} = 0 \]

Step by step solution

01

Identify the limit to be evaluated

We are given the function \(\frac{x^2}{x+1}\), and we need to find the limit as x approaches 0. The limit we need to evaluate is: \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} \]
02

Substitute \(x\) with \(0\)

To check if this limit exists, substitute \(x\) with 0 in the function: \[ \frac{0^2}{0+1} = \frac{0}{1} = 0 \] Since the function evaluates to 0, the limit exists.
03

Calculate the limit analytically

Now that we know the limit exists, we can calculate it analytically. In this case, the function is \(\frac{x^2}{x+1}\), and as x approaches 0, this simplifies to: \[ \frac{x^2}{x+1} \approx \frac{0^2}{0+1} \] \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} = \frac{0}{1} = 0 \]
04

Confirm the limit numerically

Finally, we can confirm the limit by substituting values very close to \(0\) and observing if the function continues to approach 0: For \(x = 0.1\): \[ \frac{0.1^2}{0.1+1} = \frac{0.01}{1.1} = 0.00909 \] For \(x = -0.1\): \[ \frac{-0.1^2}{-0.1+1} = \frac{0.01}{0.9} = 0.01111 \] As we can see, as the values of \(x\) get closer to \(0\), the function approaches \(0\). This confirms that the limit is indeed 0: \[ \lim_{x \rightarrow 0} \frac{x^2}{x+1} = 0 \]

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Key Concepts

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

Numerical Estimation of Limits
Numerical estimation of limits involves finding the value a function approaches as the input gets close to a certain point. It is a practical technique when analytical approaches might be complex or initially not clear.

Here is a simple step-by-step approach to estimate limits numerically:
  • Choose values of x that are increasingly close to the point of interest (in our example, 0).
  • Calculate the function value at these points.
  • Observe the trend of these function values. If they get consistently closer to a particular number, that number is the estimated limit.

When we substitute numbers like 0.1 and -0.1 into the function \(\frac{x^2}{x+1}\), our calculated values get closer and closer to 0 as x approaches 0.

Numerical methods give a good intuition of what the function behaves like around the limit, and can act as a cross-check to see if our analytical solution is on track.
Analytical Limit Calculation
Analytical limit calculation is the process of using algebraic techniques to find the limit as a variable approaches a specific value. This method provides precise and exact answers.

Some common techniques include:
  • Substitute the value directly if the function remains defined.
  • If direct substitution results in an indeterminate form like 0/0, explore simplification or factorization.
  • Utilize limit rules for algebraic functions, such as the sum, difference, product, and quotient rules.
In the given exercise, substituting 0 directly into \(\frac{x^2}{x+1}\) yielded a valid result: 0. This indicates that the function behaved nicely around that point without needing further complex calculations.

Analytical methods ensure an exact understanding of how the function behaves near the point of interest, supporting the insights gained from numerical estimation.
Limit Continuity
Limit continuity is about checking whether a function behaves "nicely" at a point, meaning that as x approaches a certain value, the function's value approaches a single, fixed number.

This concept is crucial for determining if a limit exists. A function is continuous at a point if:
  • The function is defined at that point.
  • The limit exists as the function approaches the specific point from either side.
  • The value of the function at that point matches the limit.
When evaluating \(\lim_{x \rightarrow 0} \frac{x^2}{x+1}\), we observed that the function value directly equaled the limit when 0 was substituted.

This matches the definition of a continuous function as it maintains consistency and smoothness at the point of interest, yielding an easily obtained limit.

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Most popular questions from this chapter

Annual U.S. sales of bottled water rose through the period 2000鈥�2008 as shown in the following chart The function \(R(t)=12 t^{2}+500 t+4,700\) million gallons \(\quad(0 \leq t \leq 8)\) gives a good approximation, where \(t\) is time in years since 2000 . Find the derivative function \(R^{\prime}(t)\). According to the model, how fast were annual sales of bottled water increasing in \(2005 ?\)

Your friend Muffy claims that, because the balanced difference quotient is more accurate, it would be better to use that instead of the usual difference quotient when computing the derivative algebraically. Comment on this advice.

Sketch the graph of a function whose derivative is never negative but is zero at exactly two points.

Compute the derivative function \(f^{\prime}(x)\) algebraically. (Notice that the functions are the same as those in Exercises \(1-14 .)\) HINT [See Examples 2 and \(3 .]\) $$ f(x)=-x-x^{2} $$

Compute the derivative function \(f^{\prime}(x)\) algebraically. (Notice that the functions are the same as those in Exercises \(1-14 .)\) HINT [See Examples 2 and \(3 .]\) $$ f(x)=x-2 x^{3} $$
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