Limits at infinity describe what happens to a function as \(x\) grows without bound in the positive or negative direction. Some functions approach a finite value, some grow toward infinity or negative infinity, and some do not approach one value at all. These problems include rational functions, radicals, exponential expressions, trig oscillation, and squeeze theorem examples.
Practice Problems
\(\textbf{1)}\) \(\displaystyle\lim_{x\to \infty}\frac{5x^2-10}{6x^2+1}\)
The answer is \(5/6\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5x^2-10}{6x^2+1}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{5x^2}{x^2}-\frac{10}{x^2}}{\frac{6x^2}{x^2}+\frac{1}{x^2}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5-\frac{10}{x^2}}{6+\frac{1}{x^2}}\)
\(\,\,\,\,\,\,\displaystyle\frac{5-0}{6+0}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{5}{6}\)
\(\textbf{2)}\) \(\displaystyle\lim_{x\to \infty}\frac{x^2+5x+1}{x^2+5x+100}\)
\(\)The answer is \(1\)
\(\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{x^2}{x^2}+\frac{5x}{x^2}+\frac{1}{x^2}}{\frac{x^2}{x^2}+\frac{5x}{x^2}+\frac{100}{x^2}}\)
\(\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{1+\frac{5}{x}+\frac{1}{x^2}}{1+\frac{5}{x}+\frac{100}{x^2}}\)
\(\,\,\,\,\,\displaystyle \frac{1+0+0}{1+0+0}\)
\(\,\,\,\,\,\)The answer is \(1\)
\(\textbf{3)}\) \(\displaystyle\lim_{x\to \infty}\frac{5x^2+2x-10}{3x^2+4x-5}\)
\(\)The answer is \(\displaystyle\frac{5}{3}\)
\(\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{5x^2}{x^2}+\frac{2x}{x^2}-\frac{10}{x^2}}{\frac{3x^2}{x^2}+\frac{4x}{x^2}-\frac{5}{x^2}}\)
\(\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5+\frac{2}{x}-\frac{10}{x^2}}{3+\frac{4}{x}-\frac{5}{x^2}}\)
\(\,\,\,\,\,\displaystyle \frac{5+0-0}{3+0-0}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{5}{3}\)
\(\textbf{4)}\) \(\displaystyle\lim_{x\to \infty}\frac{8x-10}{4x-5}\)
The answer is \(2\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{8x-10}{4x-5}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{8x}{x}-\frac{10}{x}}{\frac{4x}{x}-\frac{5}{x}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{8-\frac{10}{x}}{4-\frac{5}{x}}\)
\(\,\,\,\,\,\,\displaystyle\frac{8-0}{4-0}\)
\(\,\,\,\,\,\,\)The answer is \(2\)
\(\textbf{5)}\) \(\displaystyle\lim_{x\to \infty}\frac{3\sqrt{x}-4x^{1.5}}{x-2+\sqrt{x}}\)
The answer is \(-\infty\)
\(\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{3\sqrt{x}-4x^{1.5}}{x-2+\sqrt{x}}\)
\(\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{3\sqrt{x}}{x}-\frac{4x^{1.5}}{x}}{\frac{x}{x}-\frac{2}{x}+\frac{\sqrt{x}}{x}}\)
\(\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{3}{\sqrt{x}}-4\sqrt{x}}{1-\frac{2}{x}+\frac{1}{\sqrt{x}}}\)
\(\,\,\,\,\,\displaystyle\frac{0-4\sqrt{\infty}}{1-0+0}\)
\(\,\,\,\,\,\displaystyle\frac{-4\sqrt{\infty}}{1}\)
\(\,\,\,\,\,\,\)The answer is \(-\infty\)
\(\textbf{6)}\) \(\displaystyle\lim_{x\to \infty}\frac{8\sqrt{x}+x^2}{4x-x^2}\)
The answer is \(-1\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{8\sqrt{x}+x^2}{4x-x^2}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{8\sqrt{x}}{x^2}+\frac{x^2}{x^2}}{\frac{4x}{x^2}-\frac{x^2}{x^2}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{8}{x^{3/2}}+1}{\frac{4}{x}-1}\)
\(\,\,\,\,\,\,\displaystyle\frac{0+1}{0-1}\)
\(\,\,\,\,\,\,\)The answer is \(-1\)
\(\textbf{7)}\) \(\displaystyle\lim_{x\to\, -\infty}\frac{\sqrt{16x^4-3x}}{2x^2+3}\)
The answer is \(2\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{\sqrt{16x^4-3x}}{2x^2+3}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{\sqrt{x^4\left(16-\frac{3}{x^3}\right)}}{x^2\left(2+\frac{3}{x^2}\right)}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{x^2\sqrt{16-\frac{3}{x^3}}}{x^2\left(2+\frac{3}{x^2}\right)}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{\sqrt{16-\frac{3}{x^3}}}{2+\frac{3}{x^2}}\)
\(\,\,\,\,\,\,\displaystyle\frac{\sqrt{16-0}}{2+0}\)
\(\,\,\,\,\,\,\)The answer is \(2\)
\(\textbf{8)}\) \(\displaystyle\lim_{x\to\, -\infty}x^4+x^3\)
The answer is \(\infty\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\left(x^4+x^3\right)\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}x^3(x+1)\)
\(\,\,\,\,\,\,\text{As }x\to -\infty,\text{ both }x^3\text{ and }x+1\text{ are negative.}\)
\(\,\,\,\,\,\,\text{A negative times a negative becomes positive and grows without bound.}\)
\(\,\,\,\,\,\,\)The answer is \(\infty\)
\(\textbf{9)}\) \(\displaystyle\lim_{x\to \infty}\frac{4-6e^x}{2+2e^x}\)
The answer is \(-3\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{4-6e^x}{2+2e^x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{4}{e^x}-\frac{6e^x}{e^x}}{\frac{2}{e^x}+\frac{2e^x}{e^x}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{4}{e^x}-6}{\frac{2}{e^x}+2}\)
\(\,\,\,\,\,\,\displaystyle\frac{0-6}{0+2}\)
\(\,\,\,\,\,\,\)The answer is \(-3\)
\(\textbf{10)}\) \(\displaystyle\lim_{x\to \infty}e^{-x} \cos x\)
The answer is \(0\)
\(\,\,\,\,\,\,-1\le \cos{x}\le 1\)
\(\,\,\,\,\,\,-e^{-x}\le e^{-x}\cos{x}\le e^{-x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to\infty}(-e^{-x})\le \displaystyle\lim_{x\to\infty}e^{-x}\cos{x}\le \displaystyle\lim_{x\to\infty}e^{-x}\)
\(\,\,\,\,\,\,0\le \displaystyle\lim_{x\to\infty}e^{-x}\cos{x}\le 0\)
\(\,\,\,\,\,\,\)The answer is \(0\) by the Squeeze Theorem
\(\textbf{11)}\) \(\displaystyle\lim_{x\to \infty}-2 \sin x\)
The answer is Does not exist
\(\,\,\,\,\,\,-1\le \sin{x}\le 1\)
\(\,\,\,\,\,\,-2\le -2\sin{x}\le 2\)
\(\,\,\,\,\,\,\text{The function }-2\sin{x}\text{ continues to oscillate between }-2\text{ and }2.\)
\(\,\,\,\,\,\,\text{It does not approach one single value as }x\to\infty.\)
\(\,\,\,\,\,\,\)The answer is Does not exist
\(\textbf{12)}\) \(\displaystyle\lim_{x\to \infty}\frac{5x+4}{x^2-3}\)
The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5x+4}{x^2-3}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{5x}{x^2}+\frac{4}{x^2}}{\frac{x^2}{x^2}-\frac{3}{x^2}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{5}{x}+\frac{4}{x^2}}{1-\frac{3}{x^2}}\)
\(\,\,\,\,\,\,\displaystyle\frac{0+0}{1-0}\)
\(\,\,\,\,\,\,\displaystyle\frac{0}{1}\)
\(\,\,\,\,\,\,\)The answer is \(0\)
\(\textbf{13)}\) \(\displaystyle\lim_{x\to -\infty}\frac{2x-3}{7x-3}\)
The answer is \(\displaystyle\frac{2}{7}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{2x-3}{7x-3}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{\frac{2x}{x}-\frac{3}{x}}{\frac{7x}{x}-\frac{3}{x}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{2-\frac{3}{x}}{7-\frac{3}{x}}\)
\(\,\,\,\,\,\,\displaystyle\frac{2-0}{7-0}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{2}{7}\)
\(\textbf{14)}\) \(\displaystyle\lim_{x\to \infty}\frac{5x^2+4}{x^3+4x^2-3}\)
The answer is \(0\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5x^2+4}{x^3+4x^2-3}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{5x^2}{x^3}+\frac{4}{x^3}}{\frac{x^3}{x^3}+\frac{4x^2}{x^3}-\frac{3}{x^3}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{5}{x}+\frac{4}{x^3}}{1+\frac{4}{x}-\frac{3}{x^3}}\)
\(\,\,\,\,\,\,\displaystyle\frac{0+0}{1+0-0}\)
\(\,\,\,\,\,\,\)The answer is \(0\)
\(\textbf{15)}\) \(\displaystyle\lim_{x\to \infty}e^{-2x} \sin x\)
The answer is \(0\)
\(\,\,\,\,\,\,-1\le \sin{x}\le 1\)
\(\,\,\,\,\,\,-e^{-2x}\le e^{-2x}\sin{x}\le e^{-2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to\infty}(-e^{-2x})\le \displaystyle\lim_{x\to\infty}e^{-2x}\sin{x}\le \displaystyle\lim_{x\to\infty}e^{-2x}\)
\(\,\,\,\,\,\,0\le \displaystyle\lim_{x\to\infty}e^{-2x}\sin{x}\le 0\)
\(\,\,\,\,\,\,\)The answer is \(0\) by the Squeeze Theorem
\(\textbf{16)}\) \(\displaystyle\lim_{x\to \infty}e^{-2x} + \sin x\)
The answer is Does Not Exist (DNE)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to\infty}e^{-2x}=0\)
\(\,\,\,\,\,\,\sin{x}\text{ continues to oscillate between }-1\text{ and }1.\)
\(\,\,\,\,\,\,e^{-2x}+\sin{x}\text{ therefore continues to oscillate and does not approach one value.}\)
\(\,\,\,\,\,\,\)The answer is Does Not Exist (DNE)
\(\textbf{17)}\) \(\displaystyle\lim_{x\to \infty}\left(\sqrt{4x^2+3x}-2x \right)\)
The answer is \(\displaystyle\frac{3}{4}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\left(\sqrt{4x^2+3x}-2x\right)\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\left(\sqrt{4x^2+3x}-2x\right)\cdot\frac{\sqrt{4x^2+3x}+2x}{\sqrt{4x^2+3x}+2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{(4x^2+3x)-4x^2}{\sqrt{4x^2+3x}+2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{3x}{\sqrt{4x^2+3x}+2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{3}{\sqrt{4+\frac{3}{x}}+2}\)
\(\,\,\,\,\,\,\displaystyle\frac{3}{\sqrt{4+0}+2}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{3}{4}\)
\(\textbf{18)}\) \(\displaystyle\lim_{x\to -\infty}\left(\sqrt{4x^2+6x}+2x \right)\)
The answer is \(– \displaystyle\frac{3}{2}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\left(\sqrt{4x^2+6x}+2x\right)\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\left(\sqrt{4x^2+6x}+2x\right)\cdot\frac{\sqrt{4x^2+6x}-2x}{\sqrt{4x^2+6x}-2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{(4x^2+6x)-4x^2}{\sqrt{4x^2+6x}-2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{6x}{\sqrt{4x^2+6x}-2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{6x}{-x\sqrt{4+\frac{6}{x}}-2x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to -\infty}\frac{6}{-\sqrt{4+\frac{6}{x}}-2}\)
\(\,\,\,\,\,\,\displaystyle\frac{6}{-\sqrt{4+0}-2}\)
\(\,\,\,\,\,\,\)The answer is \(-\displaystyle\frac{3}{2}\)
\(\textbf{19)}\) \(\displaystyle\lim _{x\to \infty \:}\left(\frac{\sin \left(x\right)}{x}\right)\)
The answer is \(0\)
\(\,\,\,\,\,\,-1\le \sin{x}\le 1\)
\(\,\,\,\,\,\,\displaystyle-\frac{1}{x}\le \frac{\sin{x}}{x}\le \frac{1}{x}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to\infty}\left(-\frac{1}{x}\right)\le \displaystyle\lim_{x\to\infty}\frac{\sin{x}}{x}\le \displaystyle\lim_{x\to\infty}\frac{1}{x}\)
\(\,\,\,\,\,\,0\le \displaystyle\lim_{x\to\infty}\frac{\sin{x}}{x}\le 0\)
\(\,\,\,\,\,\,\)The answer is \(0\) by the Squeeze Theorem
\(\textbf{20)}\) \(\displaystyle\lim_{x\to \infty}\frac{7x^3-2x+1}{5x^3+4x^2-9}\)
The answer is \(\displaystyle\frac{7}{5}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{7x^3-2x+1}{5x^3+4x^2-9}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{\frac{7x^3}{x^3}-\frac{2x}{x^3}+\frac{1}{x^3}}{\frac{5x^3}{x^3}+\frac{4x^2}{x^3}-\frac{9}{x^3}}\)
\(\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{7-\frac{2}{x^2}+\frac{1}{x^3}}{5+\frac{4}{x}-\frac{9}{x^3}}\)
\(\,\,\,\,\,\,\displaystyle\frac{7-0+0}{5+0-0}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle\frac{7}{5}\)
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