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Chapter 2: Problem 29

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

Short Answer

Expert verified
Question: Prove that the limit of the function $$\frac{1}{(x-4)^{2}}$$ as $$x$$ approaches 4 is equal to infinity. Solution: Using the precise definition of infinite limits, we have shown that for any given $$M > 0$$, there exists a $$\delta > 0$$ such that $$0 < |x - 4| < \delta$$ implies $$\frac{1}{(x-4)^{2}} > M$$. We demonstrated that $$\delta = \sqrt{\frac{1}{M}}$$ is the appropriate value to use. Consequently, we proved that $$\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty$$.

Step by step solution

01

Write down the inequality

We want to find a $$\delta$$ such that $$0 < |x - 4| < \delta$$ implies $$\frac{1}{(x-4)^{2}} > M$$.
02

Manipulate the inequality

Since the inequality is about $$\frac{1}{(x-4)^{2}}$$, let's try to make the other side of the inequality in terms of $$\frac{1}{(x-4)^{2}}$$. We have: $$\frac{1}{(x-4)^{2}} > M$$, that is equivalent to $$\frac{1}{M}>(x-4)^2$$ since $$M>0$$.
03

Express in terms of delta

Now, we want to find a $$\delta$$ such that $$0 < |x - 4| < \delta$$ implies $$\frac{1}{M}>(x-4)^2$$. Observe that $$\delta$$ must be smaller than $$\sqrt{\frac{1}{M}}$$, since when $$0 < |x - 4| < \delta = \sqrt{\frac{1}{M}}$$, it would imply $$\frac{1}{M}>(x-4)^2$$.
04

Prove the limit

Let's show that for any given $$M > 0$$, there exists a $$\delta > 0$$ such that $$0 < |x - 4| < \delta$$ implies $$\frac{1}{(x-4)^{2}} > M$$. We know that $$\delta = \sqrt{\frac{1}{M}}$$ should work. So for $$0 < |x - 4| < \delta = \sqrt{\frac{1}{M}}$$, we have $$\frac{1}{M}>(x-4)^2$$, which implies $$\frac{1}{(x-4)^{2}} > M$$. Therefore, using the precise definition of infinite limits, we have proved that $$\lim _{x \rightarrow 4} \frac{1}{(x-4)^{2}}=\infty$$.

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