Enter a function and the calculator will determine its limits (Negative & Positive, One-tailed & Two-tailed). Get step wise solution for finite and infinite limit simplification with graph.
The limit calculator evaluates the limit of a function step-by-step as it approaches a specific value or infinity. It can analyze positive, negative, and infinity limits of any mathematical or calculus function, whether it has single or multiple variables.
“Limit defines the behavior of a function at a certain point for any input change”
Limit notation represents a mathematical concept that is based on the idea of closeness.
The calculator follows the same technique and assigns values to certain functions at points where no values are defined. It does this all in such a way as to be consistent with proximate or near values.
Limit calculator with steps works by analyzing various limit operations. These laws can be used to assess the limit of a polynomial or rational function manually as well.
Additionally, there are certain conditions for some rules and if they are not satisfied, then the rule cannot be used to validate the evaluation of a limit. Among these rules include:
Rules | Expressions |
Sum/Difference Rule | limx→b[f(x) ± h(x)] = limx→b[f(x)] ± limx→b[h(x)] |
Power Rule | limx→b[f(x)n] = [limx→b[f(x)]]n |
Product Rule | limx→b[f(x) * h(x)] = limx→b[f(x)] * limx→b[h(x)] |
Constant Rule | limx→b[k] = k |
Quotient Rule | limx→b[f(x) / h(x)] = limx→b[f(x)] / limx→b[h(x)] |
L'Hopital's Rule | limx→b[f(x) / h(x)] = limx→b[f'(x) /h'(x)] |
Evaluate the limit of the function below:
\(\lim_{x \to 3} 4x^{3}+6x{2}-x+3\)
Solution:
Here we will be using the substitution method:Apply a limit to each and every value in the given function separately to simplify the solution:
\(= \lim_{x \to 3} \left(4x^{3}\right)+\lim_{x \to 3} \left(6x^{2}\right) - \lim_{x \to 3} \left(x\right) + \lim_{x \to 3} \left(3\right)\)
Now write down each coefficient as a multiple of the separate limit functions:
\(= 4 * \lim_{x \to 3} \left(x^{3}\right)+6 * \lim_{x \to 3} \left(x^{2}\right) - \lim_{x \to 3} \left(x\right) + \lim_{x \to 3} \left(3\right)\)
Substitute the given limit i.e;
\(\lim_{x \to 3}\):
\(\lim_{x \to 3} 4x^{3}+6x{2}-x+3 = 4 * \left(3^{3}\right) + 6 * \left(3^{2}\right) - 3 + 3\)
Simplify to get the final answer:
\(\lim_{x \to 3} 4x^{3}+6x{2}-x+3 = 4 * 27 + 6 * 9 - 3 + 3\)
\(\lim_{x \to 3} 4x^{3}+6x{2}-x+3 = 108 + 6 * 9 - 3 + 3\)
\(\lim_{x \to 3} 4x^{3}+6x{2}-x+3 = 162\)
Using The Substitution Method:
\(\lim_{x \to 0} \left(\frac{sin x}{x}\right)\)
\(= \frac{sin 0}{0}\)
\(= \frac{0}{0}\)
Which is an indeterminate form. So here we will be applying l’hopital’s rule: Before we move on, we have to check whether both the functions above and below the vinculum are differentiable or not.
\(\frac{d}{dx} \left(sin x\right) = cos x\)
\(\frac{d}{dx} \left(x\right) = 1\)
Moving ahead further now:
\(\lim_{x \to 0} \left(\frac{cos x}{1}\right)\)
\(= \frac{cos 0}{1}\)
\(= 1\)
The tool is straightforward to use! It requires a few inputs to calculate limits of the given function at any point that include:
Inputs To Enter:
Results You Get:
From the authorized source of Wikipedia: Limit (mathematics), function, sequence, standard parts and much more!
The source of Khan Academy provides with: Best Strategy in finding limits