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Chapter 2: Problem 30

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

Short Answer

Expert verified
Question: Prove that the limit $$\lim_{x \to -1} \frac{1}{(x + 1)^4} = \infty$$ using the precise definition of infinite limits. Answer: To prove the limit, we've shown that for any arbitrary positive number M, there exists a 未 $$\left(未 = \sqrt[4]{\frac{1}{M}}\right)$$ such that $$\frac{1}{(x + 1)^4} > M $$ whenever $$0 < |x + 1| < 未$$. This satisfies the precise definition of infinite limits, which confirms that the limit $$\lim_{x \to -1} \frac{1}{(x + 1)^4} = \infty$$.

Step by step solution

01

To prove the limit, start with the inequality and manipulate it to find a suitable 未. Given $$\frac{1}{(x + 1)^4} > M$$, we want to find a 未 such that this inequality holds whenever $$0 < |x + 1| < 未$$. #2. Solve the inequality for x#

To find a suitable 未, we need to solve the inequality for x. Start by multiplying both sides by $$(x + 1)^4$$: $$1 > M(x + 1)^4$$. #3. Isolate x#
02

Divide both sides by M: $$\frac{1}{M} > (x + 1)^4$$. Take the fourth root of both sides: $$\sqrt[4]{\frac{1}{M}} > x + 1$$. #4. Determine 未#

Now we can see that if we choose $$未 = \sqrt[4]{\frac{1}{M}}$$, then $$0 < |x + 1| < 未$$ implies $$\frac{1}{(x + 1)^4} > M$$. #5. Choose an arbitrary M#
03

Let M be any positive number. Choose $$未 = \sqrt[4]{\frac{1}{M}}$$. #6. Apply the precise definition of infinite limits#

We've found that for any arbitrary M, there exists a 未 such that $$\frac{1}{(x + 1)^4} > M$$ whenever $$0 < |x + 1| < 未$$. This satisfies the precise definition of infinite limits, so the limit $$\lim_{x \to -1} \frac{1}{(x + 1)^4} = \infty$$.

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