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Chapter 4: Problem 54

Evaluate the following limits. $$\lim _{x \rightarrow \infty}(x-\sqrt{x^{2}+4 x})$$

Short Answer

Expert verified
Answer: The limit of the expression is 0.

Step by step solution

01

Factor out x from the square root expression

Write the given expression with x factored out of the square root: $$\lim _{x \rightarrow \infty}(x-\sqrt{x^{2}(1+\frac{4}{x})})$$
02

Simplify under the square root

Now, simplify the expression under the square root and apply the square root property: $$\lim _{x \rightarrow \infty}(x-\sqrt{x^{2}(1+\frac{4}{x})}) = \lim _{x \rightarrow \infty}(x-\sqrt{x^{2}(1+\frac{4}{x})\frac{x}{x}}) = \lim _{x \rightarrow \infty}(x-\sqrt{x^{2}\frac{x+4}{x}})$$
03

Take out the x factor

Take out the x factor from under the square root and simplify the expression: $$\lim _{x \rightarrow \infty}(x-\sqrt{x^{2}\frac{x+4}{x}}) = \lim _{x \rightarrow \infty}(x-x\sqrt{\frac{x+4}{x}})$$
04

Simplify and find the limit

Now, simplify the expression by canceling out the x: $$\lim _{x \rightarrow \infty}(x-x\sqrt{\frac{x+4}{x}}) = \lim _{x \rightarrow \infty}(x(1-\sqrt{\frac{x+4}{x}}))$$ As \(x\rightarrow\infty\), the term \(\frac{4}{x}\) approaches 0. So, $$\lim_{x\rightarrow \infty} \frac{x+4}{x} = \lim_{x\rightarrow \infty} (1+\frac{4}{x}) = 1$$ Therefore, the whole expression becomes: $$\lim _{x \rightarrow \infty}(x(1-\sqrt{\frac{x+4}{x}})) = \lim _{x \rightarrow \infty}(x(1-\sqrt{1}))$$
05

Final answer

The final answer is the simplified expression and the limit: $$\lim _{x \rightarrow \infty}(x(1-\sqrt{1}))= \lim_{x\rightarrow\infty}(x(1-1)) = \lim_{x\rightarrow\infty}(0) = 0$$

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