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Chapter 4: Problem 52

Evaluate the following limits. $$\lim _{x \rightarrow \infty}(x-\sqrt{x^{2}+1})$$

Short Answer

Expert verified
Answer: The limit is 0.

Step by step solution

01

Multiply by the conjugate expression

To simplify the expression inside the limit, we can multiply the numerator and the denominator by the conjugate expression of \(x - \sqrt{x^2 + 1}\), which is \(x + \sqrt{x^2 + 1}\). Remember to also multiply the denominator by the same expression, to preserve the overall value: $$\lim_{x \rightarrow \infty}\frac{(x-\sqrt{x^{2}+1})(x+\sqrt{x^{2}+1})}{(x+\sqrt{x^{2}+1})}$$
02

Apply the difference of squares formula

Now we can apply the difference of squares formula to the numerator, which states that \((a - b)(a + b) = a^2 - b^2\). This simplifies the numerator nicely. The limit becomes: $$\lim_{x \rightarrow \infty}\frac{x^2 - (x^2+1)}{x+\sqrt{x^{2}+1}}$$
03

Simplify the numerator

Next, we can simplify the numerator by subtracting \(x^2\) and \(-1\). The limit thus becomes: $$\lim_{x \rightarrow \infty}\frac{-1}{x+\sqrt{x^{2}+1}}$$
04

Evaluate the limit

Now, as \(x\) approaches infinity, the denominator grows without bound, since both \(x\) and \(\sqrt{x^2+1}\) are positive and grow with \(x\). As a result, the whole fraction approaches zero: $$\lim_{x \rightarrow \infty}\frac{-1}{x+\sqrt{x^{2}+1}} = 0$$ So, the given limit is equal to 0.

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