Question

Determine \(\lim f(x)\) and \(\lim _{x \rightarrow \infty} f(x)\) for the following functions. Then give the horizontal asymptote\((\text {s) of } f\) (if any). $$f(x)=\frac{\sqrt{x^{2}+1}}{2 x+1}$$

Short answer

The limit of the function as x approaches infinity is 1, and the horizontal asymptote is y = 1.

Step-by-step solution

Find the limit as x approaches infinity

To find the limit of the function as x approaches infinity, we will use the following technique: divide both the numerator and the denominator by the highest power of x in the denominator. In this case, the highest power of x in the denominator is x. So, the result will be simplified as follows: $$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\frac{\sqrt{x^{2}+1}}{x}}{\frac{2x+1}{x}}$$

Simplify the limit expression

Now, simplifying the expression becomes: $$\lim_{x \to \infty} \frac{\sqrt{x^{2}+1}\div x}{1+1\div x}$$ Now use the properties of limits: $$\lim_{x \to \infty}\frac{\sqrt{\frac{x^{2}+1}{x^2}}}{1+\frac{1}{x}}$$ Simplify further to get: $$\lim_{x \to \infty}\frac{\sqrt{1+\frac{1}{x^{2}}}}{1+\frac{1}{x}}$$

Evaluate the limit of the simplified expression

Having simplified the expression, we can evaluate the limit as x approaches infinity: $$\lim_{x \to \infty} \frac{\sqrt{1+\frac{1}{x^{2}}}}{1+\frac{1}{x}}$$ As x approaches infinity, \(\frac{1}{x^{2}}\) approaches 0 and \(\frac{1}{x}\) approaches 0, so we have: $$\lim_{x \to \infty}f(x) = \frac{\sqrt{1+0}}{1+0} = 1$$

Determine the horizontal asymptote(s)

Since the limit of the function as x approaches infinity is 1, there is a horizontal asymptote at y = 1. In standard form, the horizontal asymptote is given by the equation: $$y=1$$ #Final Answer# 1) The limit of the function as x approaches infinity is: $$\lim_{x \to \infty}f(x) = 1$$ 2) The horizontal asymptote of the function is: $$y=1$$