Limits are frequently used in calculus to find the numerical values of the functions at a specific point. After substituting the specific point, the numerical value can be finite or infinite. It is used to define the derivative calculus & integral calculus.

The complex problems of this type of calculus can be solved easily by using rules & types of limits. The power, product, quotient, & sum rules are frequently used in limits. L’hopital’s rule of limits is used if the value of the limit is undefined after applying the specific point.

In this article, we will study the basics of limits in calculus and how to solve its problems along with a lot of examples.

To find the numerical values of the function at a specific point, the limit in calculus is frequently used. It is an output of the function as the input of the function becomes closer to a specific value.

The general equation used to represent limits in calculus is:

lim_{v→a} p(v) = N

- "lim" is the notation of limits used to represent limits in calculus.
- P(v) is the function.
- "v" is the corresponding variable of limit.
- "a" is the given point.
- N is the output of the function after putting "a".

By using the above equation of limits, you can easily apply the specific point on any function to find the numerical result. Keep one thing in mind, the specific point is only applied to the corresponding variable of the function.

Below are three basic types of limits in calculus.

Right-hand limit (R-H-L) in calculus

lim_{v→a+} p(v) = N

Left-hand limit (L-H-L) in calculus

lim_{v→a-} p(v) = N

Two-sided limit (T-S-L) in calculus

lim_{v→a} p(v) = N

Below are a few general rules of limits in calculus.

Name of the rule | Rule |
---|---|

Power rule | lim_{v→a} p ^{n}(v)= [lim_{v→a}[p(v)]^{n} |

Sum rule | lim_{v→a} [p(v) * q(v)] = lim_{v→a}[p(v)] * lim_{v→a} [q(v)] |

Product rule | lim_{v→a} [p(v) * q(v)] = lim_{v→a}[p(v)] * lim_{v→a} [q(v)] |

Constant rule | lim_{v→a} F = F, where F is any constant |

Constant multiplier rule | lim_{v→a} F * p(v)= Flim_{v→a}[p(v)], where F is any constant |

Difference rule | lim_{v→a} [p(v) - q(v)] = lim_{v→a}[p(v)] -lim_{v→a} [q(v)] |

Quotient rule | lim_{v→a} [p(v) / q(v)] = lim_{v→a}[p(v)] /lim_{v→a} [q(v)] |

L'hopital's rule | lim_{v→a} [p(v) / q(v)] = lim_{v→a} [p'(v) /q'(v)] |

The problems of limits can be solved easily either by using its rules or a limit calculator. By using a calculator, you can get the

Follow the below steps to find the Solution to the given problems of limits.

- First of all, you must have a function, respective variable, & particular point.
- After identifying the given terms, write them according to the equation of limits.
- If the problem has multiple functions, use the rules of limits.
- In the end, put the specific point of limit in the obtained expression to get the result.

Below are a few examples of limits

Evaluate p(v) = 9v^{3} + 12v^{2} – 11v + 29 as “v” approaches to 3.

Step I:Write the given function, variable, & specific point. And write them according to the equation of limits.

p(v) = 9v^{3} + 12v^{2} – 11v + 29

variable of function = v

specific point = a = 3

lim_{v→a-} p(v) = lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29]

Step II:Now write the limit notation with each function separately by using the difference & sum rules of limits.

lim_{v→a-} p(v) = lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29]= lim_{v→3-} [9v^{3}] + lim_{v→3-} [12v^{2}] – lim_{v→3-} [11v] + lim_{v→3-} [29]

Step III:Use the constant multiplier rule of limits & write the constant coefficients of functions outside the limit notation.

lim_{v→a-} p(v) = lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 9lim_{v→3-} [v^{3}] + 12lim_{v→3-} [v^{2}] – 11lim_{v→3-} [v] + lim_{v→3-} [29]

Step IV:use the constant & power rules of limits & substitute v = 3 in the above expression to find the numerical result.

lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 9 [33] + 12 [32] – 11 [3] + [29]

lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 9 [3 x 3 x 3] + 12 [3 x 3] – 11 [3] + [29]

lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 9 [27] + 12 [9] – 11 [3] + [29]

lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 243 + 108 – 33 + 29

lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 351 – 33 + 29

lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 318 + 29

lim_{v→3-} [9v^{3} + 12v^{2} – 11v + 29] = 347

Evaluate p(v) = 2v^{5} + 12v^{2} + 5v - 28 as “v” approaches to 1.

Step I: Write the given function, variable, & specific point. And write them according to the equation of limits.

p(v) = 2v^{5} + 12v^{2} + 5v - 28

variable of function = v

specific point = a = 1

lim_{v→a+} p(v) = lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28]

Step II: Now write the limit notation with each function separately by using the difference & sum rules of limits.

lim_{v→a+} p(v) = lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28]= lim_{v→1+} [2v^{5}] + lim_{v→1+} [12v^{2}] + lim_{v→1+} [5v] - lim_{v→1+} [28]

Step III: Use the constant multiplier rule of limits & write the constant coefficients of functions outside the limit notation.

lim_{v→a+} p(v) = lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = 2lim_{v→1+} [v^{5}] + 12lim_{v→1+} [v^{2}] + 5lim_{v→1+} [v] - lim_{v→1+} [28]

Step IV: use the constant & power rules of limits & substitute v = 1 in the above expression to find the numerical result.

lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = 2 [15] + 12 [12] + 5 [1] - [28]

lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = 2 [1 x 1 x 1 x 1 x 1] + 12 [1 x 1] + 5 [1] - [28]

lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = 2 [1] + 12 [1] + 5 [1] - [28]

lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = 2 + 12 + 5 - 28

lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = 14 + 5 - 28

lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = 19 - 28

lim_{v→1+} [2v^{5} + 12v^{2} + 5v - 28] = -9

Evaluate p(v) = 12v^{3} – 15v^{2} + 32v + 102, when “v” approaches 2.

Step I:Write the given function, variable, & specific point. And write them according to the equation of limits.

p(v) = 12v^{3} – 15v^{2} + 32v + 102

variable of function = v

specific point = a = 2

lim_{v→a} p(v) = lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102]

Step II:Now write the limit notation with each function separately by using the difference & sum rules of limits.

lim_{v→a} p(v) = lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = lim_{v→2} [12v^{3}] – lim_{v→2} [15v^{2}] + lim_{v→2} [32v] + lim_{v→2} [102]

Step III:Use the constant multiplier rule of limits & write the constant coefficients of functions outside the limit notation.

lim_{v→a} p(v) = lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 12lim_{v→2} [v^{3}] – 15 lim_{v→2} [v^{2}] + 32 lim_{v→2} [v] + lim_{v→2} [102]

Step IV:use the constant & power rules of limits & substitute v = 2 in the above expression to find the numerical result.

lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 12[23] – 15 [22] + 32 [2] + [102]

lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 12[2 x 2 x 2] – 15 [2 x 2] + 32 [2] + [102]

lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 12[8] – 15 [4] + 32 [2] + [102]

lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 96 – 60 + 64 +102

lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 36 + 64 +102

lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 100 +102

lim_{v→2} [12v^{3} – 15v^{2} + 32v + 102] = 202

In this article, we have discussed the definition, types, rules, & solved examples of limits in calculus. Now after reading the above post, you can grab all the basics of limits in calculus. You can easily solve any complex problem of limits in calculus by following the above post.

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