Question

Limit proofs for infinite limits Use the precise definition of infinite limits to prove the following limits. $$\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty$$

Short answer

Question: Prove the following limit using the precise definition of infinite limits: $$\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty$$ Answer: In order to prove the given limit, we used the precise definition of infinite limits and found a relationship between δ and M such that if we choose δ as follows: $$\delta=\sqrt{\frac{1}{M}}$$, then the condition \(\frac{1}{(x-4)^{2}}> M\) holds true for every positive number M. Thus, we have successfully proven that $$\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty$$.

Step-by-step solution

Recall the precise definition of infinite limits

The precise definition of an infinite limit states that for a given function f(x), the limit of f(x) as x approaches a certain value is infinity if for every positive number M, there exists a positive number δ, such that if 0 < |x - a| < δ, then |f(x)| > M. In this case, we have: $$\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty$$ So we need to show that for every positive number M, there exists a positive number δ such that if 0 < |x - 4| < δ, then \(\frac{1}{(x-4)^{2}}> M\).

Find the relationship between δ and M

We want to find a δ such that if 0 < |x - 4| < δ, then \(\frac{1}{(x-4)^{2}}> M\). Let's work with the inequality: $$\frac{1}{(x-4)^{2}}> M$$ Solving for x, we obtain: $$(x-4)^{2}<\frac{1}{M}$$ Since the square root of a positive number is always positive, we can take the square root of both sides: $$|x-4|<\sqrt{\frac{1}{M}}$$ This suggests a relationship between δ and M: δ must be greater than the square root of \(\frac{1}{M}\). In other words, if we choose δ such that: $$0<|x-4| <\delta=\sqrt{\frac{1}{M}}$$ Then \(\frac{1}{(x-4)^{2}}> M\).

Prove the limit using the relationship between δ and M

Now that we have found a relationship between δ and M, let's use it to prove the given limit. We want to show that for every positive number M, there exists a positive number δ such that if 0 < |x - 4| < δ, then \(\frac{1}{(x-4)^{2}}> M\). We choose δ as follows: $$\delta=\sqrt{\frac{1}{M}}$$ For any x satisfying 0 < |x - 4| < δ, we have the following: $$\frac{1}{(x-4)^{2}}> \frac{1}{\delta^2}=\frac{1}{(\sqrt{\frac{1}{M}})^2}=M$$ Since this holds true for every positive number M, we have successfully proven that: $$\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty$$

Key concepts

Limit Proofs

Limit proofs are a fundamental aspect of calculus that provide mathematical rigor to our understanding of limits. To grapple with the idea of something approaching infinity, we need to go beyond intuitive thinking and use a precise definition to prove it. In our example, we are tasked to prove \(\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty\). This means that as \(x\) gets closer to 4, the function \(\frac{1}{(x-4)^{2}}\) grows without bound. Understanding this behavior requires evidence that for every large number \(M\), no matter how large, there is a corresponding small distance \(\delta\) near 4 such that any \(x\) within this bracket will have \(\frac{1}{(x-4)^{2}}\) greater than \(M\). This transforms theoretical mathematical behaviors into understandable, actionable concepts.

Precise Definition of Infinite Limits

The precise definition of an infinite limit is a keystone in calculus that describes what it means for a function to "tend towards infinity." This is formalized by saying that for a function \(f(x)\) to have an infinite limit as \(x\) approaches a number \(a\), we must be able to claim: for every positive number \(M\), there exists a positive delta \(\delta\) such that if \(0 < |x - a| < \delta\), then \(|f(x)| > M\).

In essence, we are ensuring that by restricting \(x\) near \(a\) with a small enough window \(\delta\), \(f(x)\) becomes fantastically large, exceeding any preselected boundary \(M\). For the problem at hand, we need to demonstrate that the function \(\frac{1}{(x-4)^{2}}\) behaves in precisely this fashion as \(x\) approaches 4. By working through this definition, we can effectively communicate the limitless growth of the function as \(x\) nears certain points.

Relationship Between δ and M

A clear comprehension of the relationship between \(\delta\) and \(M\) is crucial for proving infinite limits. When establishing \(\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty\), we need a functional interplay between these two values.

The main idea is that given any \(M\), no matter how large, there exists some \(\delta\), small but positive, such that if an \(x\)-value lies within \(\delta\) of 4, the expression \(\frac{1}{(x-4)^{2}}\) surpasses \(M\). We found this \(\delta\) using the equation \((x-4)^{2}<\frac{1}{M}\). Taking the square root, we get \(|x-4|<\sqrt{\frac{1}{M}}\), leading us to choose \(\delta = \sqrt{\frac{1}{M}}\).

This way, by shrinking \(\delta\) as \(M\) increases, we maintain the function's value exceeding \(M\), perfectly encapsulating the intrinsic infinite trait of the function within this defined range. That interaction is not just proving the limit but showcasing the powerful relationship between \(\delta\) and \(M\) that holds calculus' limits into measurable scenarios.