Limits describe what a function approaches as \(x\) gets closer and closer to a value. They are often found by direct substitution, factoring and canceling, rationalizing with a conjugate, or simplifying complex fractions. These problems include removable discontinuities, rational expressions, radical expressions, and common algebraic limit forms.
Practice Problems
\(\textbf{1)}\) \( \displaystyle \lim_{x\to3} \frac{x-3}{x^2-9} \)
The answer is \( \displaystyle \frac{1}{6} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to3} \frac{x-3}{x^2-9}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to3} \frac{x-3}{(x+3)(x-3)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to3} \frac{1}{x+3}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{3+3}\)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{1}{6} \)
\(\textbf{2)}\) \( \displaystyle \lim_{x\to5} \frac{5-x}{x^2-25} \)
The answer is \( \displaystyle -\,\frac{1}{10} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{5-x}{x^2-25} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{5-x}{(x-5)(x+5)} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{-(x-5)}{(x-5)(x+5)} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{-1}{x+5} \)
\(\,\,\,\,\,\, \displaystyle \frac{-1}{5+5} \)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle -\,\frac{1}{10} \)
\(\textbf{3)}\) \( \displaystyle \lim_{x\to 0} \frac{x^2+2x}{x} \)
The answer is \( \displaystyle 2 \)
\(\,\,\,\,\, \displaystyle \lim_{x\to 0} \frac{x^2+2x}{x}\)
\(\,\,\,\,\, \displaystyle \lim_{x\to 0} \frac{x(x+2)}{x}\)
\(\,\,\,\,\, \displaystyle \lim_{x\to 0} (x+2)\)
\(\,\,\,\,\, \displaystyle 0+2\)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle 2 \)
\(\textbf{4)}\) \( \displaystyle \lim_{x\to -1} \frac{x^2+5x+4}{x^2+4x+3} \)
The answer is \( \displaystyle \frac{3}{2} \)
\(\,\,\,\,\,\displaystyle \lim_{x\to -1} \frac{x^2+5x+4}{x^2+4x+3}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -1} \frac{(x+1)(x+4)}{(x+1)(x+3)}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -1} \frac{x+4}{x+3}\)
\(\,\,\,\,\,\displaystyle \frac{-1+4}{-1+3}\)
\(\,\,\,\,\,\displaystyle \frac{3}{2}\)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{3}{2} \)
\(\textbf{5)}\) \( \displaystyle \lim_{x\to 1} \frac{x^3-1}{x-1} \)
The hint is \(a^3-b^3=(a-b)(a^2+ab+b^2)\)
The answer is \( 3 \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to 1} \frac{x^3-1}{x-1} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{x-1} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to 1} (x^2+x+1) \)
\(\,\,\,\,\,\, \displaystyle (1)^2+(1)+1 \)
\(\,\,\,\,\,\,\)The answer is \( 3 \)
\(\textbf{6)}\) \( \displaystyle \lim_{x\to1} \frac{x^2-1}{x^3-1} \)
The hint is \(a^3-b^3=(a-b)(a^2+ab+b^2)\)
The answer is \( \displaystyle \frac{2}{3} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to1} \frac{x^2-1}{x^3-1} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to1} \frac{(x-1)(x+1)}{(x-1)(x^2+x+1)} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to1} \frac{x+1}{x^2+x+1} \)
\(\,\,\,\,\,\, \displaystyle \frac{1+1}{1^2+1+1} \)
\(\,\,\,\,\,\, \displaystyle \frac{2}{3} \)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{2}{3} \)
\(\textbf{7)}\) \( \displaystyle \lim_{x\to 4} \frac{4x-x^2}{2-\sqrt{x}} \)
The hint is \( \left(4-x\right)=\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right) \)
The answer is \( 16 \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} \frac{4x-x^2}{2-\sqrt{x}} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} \frac{x(4-x)}{2-\sqrt{x}} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} \frac{x(2-\sqrt{x})(2+\sqrt{x})}{2-\sqrt{x}} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} x(2+\sqrt{x}) \)
\(\,\,\,\,\,\, \displaystyle (4)(2+\sqrt{4}) \)
\(\,\,\,\,\,\, \displaystyle (4)(2+2) \)
\(\,\,\,\,\,\,\) The answer is \( 16 \)
\(\textbf{8)}\) \( \displaystyle \lim_{x\to 3} \frac{3x}{x^2+2} \)
The answer is \( \displaystyle \frac{9}{11} \)
\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{3x}{x^2+2}\)
\(\,\,\,\,\, \displaystyle \frac{3(3)}{(3)^2+2}\)
\(\,\,\,\,\, \displaystyle \frac{9}{9+2}\)
\(\,\,\,\,\, \displaystyle \frac{9}{11}\)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{9}{11} \)
\(\textbf{9)}\) \( \displaystyle \lim_{x\to 3} \frac{\frac{1}{x}-\frac{1}{3}}{x-3} \)
The answer is \( \displaystyle -\frac{1}{9} \)
\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{\frac{1}{x}-\frac{1}{3}}{x-3}\)
\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{\frac{3-x}{3x}}{x-3}\)
\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{-(x-3)}{3x(x-3)}\)
\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{-1}{3x}\)
\(\,\,\,\,\, \displaystyle \frac{-1}{3(3)}\)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle -\frac{1}{9} \)
\(\textbf{10)}\) \( \displaystyle \lim_{x\to 0} \frac{1-\sqrt{1-x}}{x} \)
The answer is \( \displaystyle \frac{1}{2} \)
\(\,\,\,\,\,\displaystyle \lim_{x\to 0} \frac{1-\sqrt{1-x}}{x}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to 0} \frac{1-\sqrt{1-x}}{x}\cdot\frac{1+\sqrt{1-x}}{1+\sqrt{1-x}}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to 0} \frac{1-(1-x)}{x(1+\sqrt{1-x})}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to 0} \frac{x}{x(1+\sqrt{1-x})}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to 0} \frac{1}{1+\sqrt{1-x}}\)
\(\,\,\,\,\,\displaystyle \frac{1}{1+\sqrt{1}}\)
\(\,\,\,\,\,\,\)The answer is \(\displaystyle \frac{1}{2}\)
\(\textbf{11)}\) \( \displaystyle \lim_{x\to3} \frac{x^2-1}{x^3-1} \)
The answer is \( \displaystyle \frac{4}{13} \)
\(\,\,\,\,\,\, \displaystyle \lim_{x\to3} \frac{x^2-1}{x^3-1} \)
\(\,\,\,\,\,\, \displaystyle \frac{(3)^2-1}{(3)^3-1} \)
\(\,\,\,\,\,\, \displaystyle \frac{9-1}{27-1} \)
\(\,\,\,\,\,\, \displaystyle \frac{8}{26} \)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{4}{13} \)
\(\textbf{12)}\) \( \displaystyle \lim_{x\to 4} \frac{\sqrt{x} – 2}{x – 4} \)
The answer is \( \displaystyle \frac{1}{4} \)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to 4} \frac{\sqrt{x} – 2}{x – 4}\)
\(\,\,\,\,\,\,\text{Multiply numerator and denominator by the conjugate.}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to 4} \frac{(\sqrt{x} – 2)(\sqrt{x} + 2)}{(x – 4)(\sqrt{x} + 2)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to 4} \frac{x – 4}{(x – 4)(\sqrt{x} + 2)}\)
\(\,\,\,\,\,\,\displaystyle \lim_{x\to 4} \frac{1}{\sqrt{x} + 2}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{\sqrt{4} + 2}\)
\(\,\,\,\,\,\,\displaystyle \frac{1}{2 + 2}\)
\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{1}{4} \)
\(\textbf{13)}\) \(\displaystyle \lim_{x\to2}\frac{x^2-4}{x-2}\)
The answer is \(4\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2}\frac{x^2-4}{x-2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2}\frac{(x-2)(x+2)}{x-2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2}(x+2)\)
\(\,\,\,\,\,2+2\)
\(\,\,\,\,\,\)The answer is \(4\)
\(\textbf{14)}\) \(\displaystyle \lim_{x\to -2}\frac{x^2+x-2}{x+2}\)
The answer is \(-3\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -2}\frac{x^2+x-2}{x+2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -2}\frac{(x+2)(x-1)}{x+2}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -2}(x-1)\)
\(\,\,\,\,\,-2-1\)
\(\,\,\,\,\,\)The answer is \(-3\)
\(\textbf{15)}\) \(\displaystyle \lim_{x\to0}\frac{\sqrt{x+9}-3}{x}\)
The answer is \(\displaystyle\frac{1}{6}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sqrt{x+9}-3}{x}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sqrt{x+9}-3}{x}\cdot\frac{\sqrt{x+9}+3}{\sqrt{x+9}+3}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{x+9-9}{x(\sqrt{x+9}+3)}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{x}{x(\sqrt{x+9}+3)}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{1}{\sqrt{x+9}+3}\)
\(\,\,\,\,\,\displaystyle \frac{1}{\sqrt{9}+3}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle\frac{1}{6}\)
\(\textbf{16)}\) \(\displaystyle \lim_{x\to1}\frac{x^2-2x+1}{x-1}\)
The answer is \(0\)
\(\,\,\,\,\,\displaystyle \lim_{x\to1}\frac{x^2-2x+1}{x-1}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to1}\frac{(x-1)^2}{x-1}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to1}(x-1)\)
\(\,\,\,\,\,1-1\)
\(\,\,\,\,\,\)The answer is \(0\)
\(\textbf{17)}\) \(\displaystyle \lim_{x\to -3}\frac{x^2+6x+9}{x+3}\)
The answer is \(0\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -3}\frac{x^2+6x+9}{x+3}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -3}\frac{(x+3)^2}{x+3}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to -3}(x+3)\)
\(\,\,\,\,\,-3+3\)
\(\,\,\,\,\,\)The answer is \(0\)
\(\textbf{18)}\) \(\displaystyle \lim_{x\to0}\frac{\sin x}{x}\)
The answer is \(1\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin x}{x}\)
\(\,\,\,\,\,\text{This is one of the basic trig limits.}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1\)
\(\,\,\,\,\,\)The answer is \(1\)
\(\textbf{19)}\) \(\displaystyle \lim_{x\to0}\frac{\tan x}{x}\)
The answer is \(1\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\tan x}{x}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin x}{x\cos x}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to0}\frac{\sin x}{x}\cdot\frac{1}{\cos x}\)
\(\,\,\,\,\,\displaystyle (1)\cdot\frac{1}{1}\)
\(\,\,\,\,\,\)The answer is \(1\)
\(\textbf{20)}\) \(\displaystyle \lim_{x\to2}\frac{x^2-5x+6}{x^2-4}\)
The answer is \(\displaystyle -\frac{1}{4}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2}\frac{x^2-5x+6}{x^2-4}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2}\frac{(x-2)(x-3)}{(x-2)(x+2)}\)
\(\,\,\,\,\,\displaystyle \lim_{x\to2}\frac{x-3}{x+2}\)
\(\,\,\,\,\,\displaystyle \frac{2-3}{2+2}\)
\(\,\,\,\,\,\)The answer is \(\displaystyle -\frac{1}{4}\)
See Related Pages\(\)
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