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Problem 321

Find $\lim _{x \rightarrow \infty}\left[x \sqrt{\left(x^{2}+1\right)}-x^{2}\right]$

Short Answer

Expert verified
The short answer to the limit question is: \(\lim_{x \rightarrow \infty}\left[x\sqrt{x^2+1}-x^{2}\right] = 0\).
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Step by step solution

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Step 1: Simplify the Expression inside the Square Root

First, let's simplify the expression inside the square root: \(\sqrt{x^2+1}\) Notice that the leading term of the polynomial is x^2, so we can factor this out of the expression: \(\sqrt{x^2(1 + \frac{1}{x^2})}\) Now, the expression inside the square root simplifies to: \(\sqrt{x^2} \cdot \sqrt{1 + \frac{1}{x^2}} = x \cdot \sqrt{1 + \frac{1}{x^2}}\) Our expression becomes: \(\lim_{x \rightarrow \infty}\left[x(x \cdot \sqrt{1 + \frac{1}{x^2}})-x^{2}\right]\)

Step 2: Factor out the x Term

We can factor out the x term from the expression: \(\lim _{x \rightarrow \infty}\left[x\sqrt{1 + \frac{1}{x^2}} \cdot x-x^{2}\right] = \lim _{x \rightarrow \infty} x\left(\sqrt{1 + \frac{1}{x^2}} - 1\right)\)

Step 3: Substitute u for 1/x&

In order to simplify the expression further, we can let \(u = \frac{1}{x}\) (since x approaches infinity, u approaches 0) Now, our expression becomes: \(\lim_{u \rightarrow 0} \frac{1}{u}\left(\sqrt{1 + u^2} - 1\right)\)

Step 4: Rationalize the Denominator

In order to deal with the square root, we can rationalize the denominator: Multiply both the numerator and the denominator by \(\sqrt{1 + u^2} + 1\) to get: \(\lim_{u \rightarrow 0} \frac{(\sqrt{1 + u^2} - 1)(\sqrt{1 + u^2} + 1)}{u(\sqrt{1 + u^2} + 1)}\)

Step 5: Simplify the Expression

Now, we can simplify the expression by multiplying the numerator and denominator expressions: \(\lim_{u \rightarrow 0} \frac{(1 + u^2) - 1}{u(\sqrt{1 + u^2} + 1)} = \lim_{u \rightarrow 0} \frac{u^2}{u(\sqrt{1 + u^2} + 1)}\) Now, we can cancel out the u term in the numerator and denominator: \(\lim_{u \rightarrow 0} \frac{u}{\sqrt{1 + u^2} + 1}\)

Step 6: Find the Limit as u Approaches 0

Now, we can find the limit as u approaches 0: \(\lim_{u \rightarrow 0} \frac{u}{\sqrt{1 + u^2} + 1} = \frac{0}{\sqrt{1 + 0^2} + 1} = \frac{0}{1 + 1} = 0\) So, the limit of the given expression as x approaches infinity is 0.

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Most popular questions from this chapter

Chapter 11

Show that the following sequences are convergent: a) $a_{n}=[\\{1 \cdot 3 \cdot 5 \ldots(2 n-1)\\} /\\{2 \cdot 4 \cdot 6 \ldots(2 n)\\}]$ \(\mathrm{n}=1,2,3, \ldots\) b) \(a_{n}=(1 / 1 !)+(1 / 2 !)+\ldots+(1 / n !)\) \(\mathrm{n}=1,2,3, \ldots\)

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Show that: a) $\lim _{n \rightarrow \infty}\left[\left\\{n^{4}+n^{3}-1\right\\} /\left\\{\left(n^{2}+2\right)\left(n^{2}-n-1\right)\right\\}\right]=1$ b) $\lim _{n \rightarrow \infty}\left[1+\left(\mathrm{C} / \mathrm{n}^{2}\right)\right]^{\mathrm{n}}=1$, C a constant.

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Prove: a) The sequence of functions \(\left\\{\mathrm{f}_{\mathrm{n}}\right\\}\) defined on \(\mathrm{E}\), converges uniformly on \(E\) if and only if for every \(\varepsilon>0\) there exists an integer \(\mathrm{N}\) such that $\mathrm{m} \geq \mathrm{N}, \mathrm{n} \geq \mathrm{N}, \mathrm{x} \in \mathrm{E}$ implies $1 \mathrm{f}_{\mathrm{n}}(\mathrm{x})-\mathrm{f}_{\mathrm{m}}(\mathrm{x}) \mid \leq \varepsilon$ b) Suppose \(\left\\{f_{n}\right\\}\) is a sequence of functions defined on \(E\), and suppose $1 \mathrm{f}_{\mathrm{n}}(\mathrm{x}) 1 \leq \mathrm{M}_{\mathrm{n}}$ \((\mathrm{x} \in \mathrm{E}, \mathrm{n}=1,2,3, \ldots)\), where $\mathrm{M}_{\mathrm{n}}=\sup _{\mathrm{x} \in \mathrm{E}}\left|\mathrm{f}_{\mathrm{n}}(\mathrm{x})\right|$ and $\lim _{\mathrm{n} \rightarrow \infty} \mathrm{f}_{\mathrm{n}}(\mathrm{x})=0, \mathrm{x} \in \mathrm{E}$, then \({ }^{\infty} \sum_{n=1} f_{n}\) converges uniformly on \(\mathrm{E}\) if \(\sum \mathrm{M}_{\mathrm{n}}\) converges.

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