L'hopital's Rule Calculator with steps
L'hopital's Rule Calculator is used to find the limits of the undefined functions. This calculator takes the derivatives of the undefined function and put the limit value to get the numerical result.
How does this L'hopital calculator work?
Follow the below steps to find the limits of function using L'hopital's rule.
- Input the function.
- Use the keypad icon
to enter mathematical keys. - Enter the limit value and select the variable.
- Select left-hand, right-hand, or two-sided limit.
- Click the calculate button.
- To enter a new function, press the reset button.
- Hit the show more button to view the result with steps.
What is L'hopital's rule?
In calculus, L'hopital's rule is a theorem of limits that helps us to calculate undefined limits of the form of \(\frac{0}{0}\:or\:\frac{\infty }{\infty }\)
In simple words, L'hopital's rule helps us to find the \(\lim _{x\to a}\left(\frac{g\left(x\right)}{h\left(x\right)}\right)\:\)
Where \(\lim _{x\to a}\:g\left(x\right)=\lim _{x\to a}\:h\left(x\right)=0\:or\:\left(\infty \:,-\infty \right)\)
The formula of L'hopital's rule
According to this rule, if the derivatives of the functions exist then two limits are equivalent. The general formula of this rule is given below.
\(\lim _{x\to a}\left(\frac{g\left(x\right)}{h\left(x\right)}\right)=\lim _{x\to a}\left(\frac{g'\left(x\right)}{h'\left(x\right)}\right)\)
How to use L'hopital's rule to find the limits?
Following is an example of this rule solved by our L'hospital calculator.
Example 1
Evaluate \(\lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)\).
Solution
Step 1: Apply the limit value and put 0 in the place of x.
\(\lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)=\frac{sin\left(0\right)}{0}\)
\(\lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)=\frac{0}{0}\)
Step 2: Use the L'hopital's rule as the given function gives \(\frac{0}{0}\) form.
\(\lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)=\lim \:_{x\to \:0}\left(\frac{\frac{d}{dx}sin\left(x\right)}{\frac{d}{dx}x}\right)\)
\(\lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)=\lim _{x\to 0}\left(\frac{cos\left(x\right)}{1}\right)\)
\(\lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)=\lim _{x\to 0}\left(cos\left(x\right)\right)\)
\(\lim _{x\to 0}\left(\frac{sin\left(x\right)}{x}\right)=cos\left(0\right)=1\)
Example 2
Evaluate \(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)\)
Solution
Step 1: Substitute the limit value in the function.
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\frac{9\left(\infty \right)+21}{3\left(\infty \right)+4}\)
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\frac{\infty +21}{\infty \:+4}=\frac{\infty }{\infty }\)
Step 2: Use the L'hopistal's rule to find the limit of \(\frac{\infty }{\infty }\) function.
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\lim _{x\to \infty \:}\left(\frac{\frac{d}{dx}\left(9x+21\right)}{\frac{d}{dx}\left(3x^2+4\right)}\right)\)
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\lim _{x\to \infty \:}\left(\frac{\frac{d}{dx}9x+\frac{d}{dx}21}{\frac{d}{dx}3x^2+\frac{d}{dx}4}\right)\)
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\lim _{x\to \infty \:}\left(\frac{9+0}{6x+0}\right)\)
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\lim _{x\to \infty \:}\left(\frac{9}{6x}\right)\)
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\frac{9}{6\left(\infty \right)}\)
\(\lim _{x\to \infty }\left(\frac{9x+21}{3x^2+4}\right)=\frac{9}{\infty }=0\)
References
Khan Academy. (n.d.). What is L'Hôpital's Rule? . Khan Academy.
Examples of L'hopital's rule | L'Hopital's rule. (n.d.).
