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\(\textbf{1)}\) \( \displaystyle \lim_{x\to3} \frac{x-3}{x^2-9} \)

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The answer is \( \displaystyle \frac{1}{6} \)

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\(\,\,\,\,\,\,\displaystyle \lim_{x\to3} \frac{x-3}{x^2-9}\)

\(\,\,\,\,\,\,\displaystyle \lim_{x\to3} \frac{x-3}{(x+3)(x-3)}\)

\(\,\,\,\,\,\,\displaystyle \lim_{x\to3} \frac{1}{(x+3)}\)

\(\,\,\,\,\,\,\displaystyle \frac{1}{((3)+3)}\)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{1}{6} \)

\(\textbf{2)}\) \( \displaystyle \lim_{x\to5} \frac{5-x}{x^2-25} \)

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The answer is \( \displaystyle -\,\frac{1}{10} \)

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\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{5-x}{x^2-25} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{5-x}{(x-5)(x+5)} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{5-x}{-1 \cdot (5-x)(x+5)} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to5} \frac{1}{-1 \cdot (x+5)} \)

\(\,\,\,\,\,\, \displaystyle \frac{1}{-1 \cdot (5+5)} \)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle -\,\frac{1}{10} \)

\(\textbf{3)}\) \( \displaystyle \lim_{x\to 0} \frac{x^2+2x}{x} \)

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The answer is \( \displaystyle 2 \)

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\(\,\,\,\,\, \displaystyle \lim_{x\to 0} \frac{x^2+2x}{x}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 0} \frac{x(x+2)}{x}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 0} \frac{(x+2)}{1}\)

\(\,\,\,\,\, \displaystyle \frac{(0+2)}{1}\)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle 2 \)

\(\textbf{4)}\) \( \displaystyle \lim_{x\to -1} \frac{x^2+5x+4}{x^2+4x+3} \)

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The answer is \( \displaystyle \frac{3}{2} \)

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\(\,\,\,\,\,\displaystyle \lim_{x\to -1} \frac{(x+1)(x+4)}{(x+1)(x+3)}\)

\(\,\,\,\,\,\displaystyle \lim_{x\to -1} \frac{(x+4)}{(x+3)}\)

\(\,\,\,\,\,\displaystyle \frac{(-1)+4}{(-1)+3}\)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{3}{2} \)

\(\textbf{5)}\) \( \displaystyle \lim_{x\to 1} \frac{x^3-1}{x-1} \)

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The hint is \(a^3-b^3=(a-b)(a^2+ab+b^2)\)

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The answer is \( 3 \)

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\(\,\,\,\,\,\, \displaystyle \lim_{x\to 1} \frac{x^3-1}{x-1} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{x-1} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to 1} \frac{(x^2+x+1)}{1} \)

\(\,\,\,\,\,\, \displaystyle \frac{((1)^2+(1)+1)}{1} \)

\(\,\,\,\,\,\,\)The answer is \( 3 \)

\(\textbf{6)}\) \( \displaystyle \lim_{x\to1} \frac{x^2-1}{x^3-1} \)

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The hint is \(a^3-b^3=(a-b)(a^2+ab+b^2)\)

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The answer is \( \displaystyle \frac{2}{3} \)

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\(\,\,\,\,\,\, \displaystyle \lim_{x\to1} \frac{x^2-1}{x^3-1} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to1} \frac{(x-1)(x+1)}{(x-1)(x^2+x+1)} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to1} \frac{(x+1)}{(x^2+x+1)} \)

\(\,\,\,\,\,\, \displaystyle \frac{((1)+1)}{((1)^2+(1)+1)} \)

\(\,\,\,\,\,\, \displaystyle \frac{(2)}{(3)} \)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{2}{3} \)

\(\textbf{7)}\) \( \displaystyle \lim_{x\to 4} \frac{4x-x^2}{2-\sqrt{x}} \)

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The hint is \( \left(4-x\right)=\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right) \)

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The answer is \( 16 \)

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\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} \frac{4x-x^2}{2-\sqrt{x}} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} \frac{x\left(4-x\right)}{2-\sqrt{x}} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} \frac{x\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}{2-\sqrt{x}} \)

\(\,\,\,\,\,\, \displaystyle \lim_{x\to 4} \frac{x\left(2+\sqrt{x}\right)}{1} \)

\(\,\,\,\,\,\, \displaystyle \frac{(4)\left(2+\sqrt{4}\right)}{1} \)

\(\,\,\,\,\,\, \displaystyle \frac{(4)\left(2+2\right)}{1} \)

\(\,\,\,\,\,\,\) The answer is \( 16 \)

\(\textbf{8)}\) \( \displaystyle \lim_{x\to 3} \frac{3x}{x^2+2} \)

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The answer is \( \displaystyle \frac{9}{11} \)

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\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{3x}{x^2+2}\)

\(\,\,\,\,\, \displaystyle \frac{3(3)}{(3)^2+2}\)

\(\,\,\,\,\, \displaystyle \frac{9}{9+2}\)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{9}{11} \)

\(\textbf{9)}\) \( \displaystyle \lim_{x\to 3} \frac{\frac{1}{x}-\frac{1}{3}}{x-3} \)

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The answer is \( \displaystyle -\frac{1}{9} \)

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\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{\frac{1}{x}-\frac{1}{3}}{x-3}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{\frac{3}{3x}-\frac{x}{3x}}{\frac{x-3}{1}}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{\frac{3-x}{3x}}{\frac{x-3}{1}}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{3-x}{3x} \cdot \frac{1}{x-3}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{-1\left(x-3\right)}{3x} \cdot \frac{1}{x-3}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{-1}{3x} \cdot \frac{1}{1}\)

\(\,\,\,\,\, \displaystyle \lim_{x\to 3} \frac{-1}{3x}\)

\(\,\,\,\,\, \displaystyle \frac{-1}{3(3)}\)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle -\frac{1}{9} \)

\(\textbf{10)}\) \( \displaystyle \lim_{x\to 0} \frac{1-\sqrt{1-x}}{x} \)

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The answer is \( \displaystyle \frac{1}{2} \)

\(\textbf{11)}\) \( \displaystyle \lim_{x\to3} \frac{x^2-1}{x^3-1} \)

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The answer is \( \displaystyle \frac{4}{13} \)

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\(\,\,\,\,\,\, \displaystyle \lim_{x\to3} \frac{x^2-1}{x^3-1} \)

\(\,\,\,\,\,\, \displaystyle \frac{(3)^2-1}{(3)^3-1} \)

\(\,\,\,\,\,\, \displaystyle \frac{9-1}{27-1} \)

\(\,\,\,\,\,\, \displaystyle \frac{8}{26} \)

\(\,\,\,\,\,\,\)The answer is \( \displaystyle \frac{4}{13} \)

See Related Pages\(\)

\(\bullet\text{ Limit Calculator}\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Limits on Graphs}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Continuity on Graphs}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Piecewise Functions- Limits and Continuity}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Limits at Infinity}\)
\(\,\,\,\,\,\,\,\,\displaystyle\lim_{x\to \infty}\frac{5x^2+2x-10}{3x^2+4x-5}…\)

\(\bullet\text{ Trig Limits}\)
\(\,\,\,\,\,\,\,\,\displaystyle \lim_{\theta\to0} \frac{\sin \theta}{\theta}=1…\)

In Summary

Limits is a fundamental concept in calculus that deals with the behavior of functions as they approach certain values. It allows us to understand and predict the behavior of functions as they approach infinity or as certain variables within the function approach certain values.

The definition of limits is a bit technical, but it can be thought of as a way to determine the value of a function at a specific point, even if the function itself is not defined at that point. This is done by considering the values of the function as it approaches the specific point from both sides.

We learn about limits because it is an essential concept in calculus and has numerous applications in fields such as physics, engineering, and economics. It allows us to understand and analyze the behavior of functions in a way that would not be possible without it.

Limits are typically introduced in Calculus courses.

Common mistakes with limits usually occur when evaluating limits that involve infinite or indeterminate forms.

Limits were first developed and formalized by Newton and Leibniz in the 17th century.

Math topics that use limits include derivatives, which are used to measure the rate of change of a function, and integrals, which allow us to find the area under a curve. Both of these concepts are also essential in calculus.

Some math topics that use intro to limits

Derivatives: Limits are a key concept in the definition of the derivative, which is a measure of the instantaneous slope of a function at a particular point. This is found by taking the limit as the change in x approaches zero.

Continuity: A function is continuous at a point if, roughly speaking, it does not have any “breaks” or “jumps” at that point. This idea can be made precise using limits: a function is continuous at a point if the limit of the function as the input approaches that point from both sides is equal to the value of the function at that point.

Integration: Integration is a way of finding the area under a curve or the volume of a three-dimensional object by dividing it into smaller pieces and adding them up. One way to do this is by using limits, which allow us to calculate the area or volume of the object by taking the limit as the size of the pieces approaches zero.

Asymptotes: An asymptote is a line that a curve approaches as it extends indefinitely in some direction. For example, the graph of the function \(y = \frac{1}{x^2}\) has a vertical asymptotes at x = 0, because the function becomes arbitrarily large as x approaches 0 from the left or right. Asymptotes can be analyzed using limits.

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