What is L’Hospital’s rule? In calculus, L’Hospital’s rule is a powerful tool to evaluate the limits of indeterminate forms. This rule shows whether a limit exists or not; if yes, then we can determine its exact value. In short, this rule tells us that in case we have indeterminate forms, like 0/0 and ∞/∞ , then we just differentiate the numerator, as well as the denominator and simplify the evaluation of limits. This article helps students to learn how to use L’Hospital’s rule for solving indeterminate limits.
What Do You Mean by Indeterminate Forms?
The indeterminate form is a mathematical expression that we cannot determine the original value even after applying the limits.
Suppose we have to calculate a limit of f(x) at x→a. Then, we first check whether it is an indeterminate form or not by directly putting the value of x=a in the given function. If we get f(a)/g(a) = 0/0, ∞/∞, 0 . ∞/∞, 1∞, ∞/∞, these all are called indeterminate forms. L’Hospital’s rule is applicable in the first two cases, i.e., 0/0 and ∞/∞.
L’Hospital’s Rule Formula
Solved Problems
Example 1: Find \(\begin{array}{l}\lim_{x\to 3} \frac{x^{2}-x-6}{x-3}\end{array} \)
Solution:
Given \(\begin{array}{l}\lim_{x\to 3} \frac{x^{2}-x-6}{x-3}\end{array} \)
If we put x = 3, we get a solution of the form 0/0.
Apply L’hospital rule.
Differentiate the numerator and denominator, and we get,
\(\begin{array}{l}\lim_{x\to 3} \frac{x^{2}-x-6}{x-3}\end{array} \)
\(\begin{array}{l}=\lim_{x\to 3} \frac{2x-1}{1}\end{array} \)
= [2(3)-1]/1 = (6-1)/1 [substitute x = 3]
= 5
Hence, the required solution is 5.
Example 2: Solve
\(\begin{array}{l}lim_{x \rightarrow 0} \frac{sin\ x }{x}\end{array} \)
Solution:
Put x = 0 in the given function, and we get 0/0.
To solve this question, we differentiate the numerator and denominator separately.
\(\begin{array}{l}lim_{x \rightarrow 0} \frac{cos\ x }{1}\end{array} \)
= cos(0)/1 = 1/1 = 1
Example 3: Solve
\(\begin{array}{l}lim_{x \rightarrow 0} \frac{e^x – cos\ x }{sin\ 2x}\end{array} \)
Solution:
At x = 0, the given function is in 0/0 form.
On differentiating the given function, we get
\(\begin{array}{l}lim_{x \rightarrow 0} \frac{e^x – sin\ x }{2\ cos\ 2x}\end{array} \)
= 1/2
Example 4: Find the limit for the below functions.
\(\begin{array}{l}(i)\ lim_{x \rightarrow \infty} \frac{4x^2-5x}{3x^2-1}\\ (ii)\ lim_{x \rightarrow 1} \frac{5x^4-4x^2-1}{9x^3+x-10}\\ (iii)\ lim_{x \rightarrow 0^+} \frac{e^{3x}}{5x+200}\\\end{array} \)
Solution:
(i) At x = ∞, the given function is in ∞/∞ form.
On differentiating the given function twice, we have
\(\begin{array}{l}lim_{x \rightarrow \infty} \frac{4x^2-5x}{3x^2-1}\end{array} \)
\(\begin{array}{l}=lim_{x \rightarrow \infty}\end{array} \)
8/6 = 4/3
(ii) At x = 1, the given function is in 0/0 form.
Differentiate the given function.
\(\begin{array}{l}lim_{x \rightarrow 1} \frac{20x^3-8x}{27x^2+1}\end{array} \)
= 12/28 = 3/7
(iii)
\(\begin{array}{l}lim_{x \rightarrow 0^+} \frac{3e^{3x}}{5}\end{array} \)
= 3/5
Example 5: Evaluate
\(\begin{array}{l}\lim_{x \rightarrow 0^+} \frac{x^{-1/3}}{ln(x)} \end{array} \)
Solution:
\(\begin{array}{l}\lim_{x \rightarrow 0^+} \frac{x^{-1/3}}{ln(x)} \end{array} \)
Example 6: Evaluate limit for the function,
\(\begin{array}{l}\lim_{x\rightarrow 0^+}\frac{sin2x}{\sqrt{5x}}\end{array} \)
Solution:
Example 7: Solve the limit for the function applying L’Hospital’s rule.
\(\begin{array}{l}\lim_{x\rightarrow 0^{-}}\frac{1-cost}{t^{3}}\end{array} \)
Solution:
\(\begin{array}{l}\lim_{x\rightarrow 0^{-}}\frac{1-cost}{t^{3}} = \frac{1-cos0}{0}=\frac{0}{0}\end{array} \)
(Indeterminate form)
Applying L’Hospital’s rule, we have
\(\begin{array}{l}\lim_{x\rightarrow 0^{-}}\frac{1-cost}{t^{3}}\\=\lim _{t\to \:0-}\left(\frac{\sin \left(t\right)}{3t^2}\right)\\=\lim _{t\to \:0-}\left(\frac{\cos \left(t\right)}{6t}\right)\\=\frac{1}{6}\cdot \lim _{t\to \:0-}\left(\frac{\cos \left(t\right)}{t}\right)\\=\frac{1}{6}\cdot \lim _{t\to \:0-}\left(\cos \left(t\right)\frac{1}{t}\right)\\=\frac{1}{6}\cdot \lim _{t\to \:0-}\left(\cos \left(t\right)\right)\cdot \lim _{t\to \:0-}\left(\frac{1}{t}\right)\\\lim _{t\to \:0-}\left(\cos \left(t\right)\right)\\=\cos \left(0\right)\\=1\\\lim _{t\to \:0-}\left(\frac{1}{t}\right)\\=-\infty \:\\=\frac{1}{6}\cdot \:1\cdot \left(-\infty \:\right)\\=-\infty \:\\\end{array} \)
Example 8: Evaluate the limit
\(\begin{array}{l}\lim _{x\to \:1-}\left(\frac{1-x+\ln \left(x\right)}{1+\cos \left(\pi x\right)}\right)\end{array} \)
Solution:
\(\begin{array}{l}\lim _{x\to \:1-}\left(\frac{1-x+\ln \left(x\right)}{1+\cos \left(\pi x\right)}\right)\end{array} \)
Apply L’Hospital’s rule,
\(\begin{array}{l}\lim _{x\to \:1-}\left(\frac{1-x+\ln \left(x\right)}{1+\cos \left(\pi x\right)}\right)\\=\lim _{x\to \:1-}\left(\frac{\frac{1}{x}-1}{-\pi \sin \left(\pi x\right)}\right)\\=-\frac{\frac{1}{x}-1}{\pi \sin \left(\pi x\right)}\\=-\frac{\frac{-x+1}{x}}{\pi \sin \left(\pi x\right)}\\=-\frac{-x+1}{\pi x\sin \left(\pi x\right)}\\=\lim _{x\to \:1-}\left(-\frac{1-x}{\pi x\sin \left(\pi x\right)}\right)\\=\lim _{x\to \:1-}\left(\frac{1}{\pi \left(\sin \left(\pi x\right)+\pi x\cos \left(\pi x\right)\right)}\right)\\=\frac{1}{\pi \left(\sin \left(\pi 1\right)+\pi 1\cdot \cos \left(\pi 1\right)\right)}\\\pi \left(\sin \left(\pi 1\right)+\pi 1\cdot \cos \left(\pi 1\right)\right)=-\pi ^2\\=\frac{1}{-\pi ^2}\\=-\frac{1}{\pi ^2}\\\end{array} \)
Example 9: Evaluate
\(\begin{array}{l}\lim _{x\rightarrow \:0}\frac{\sqrt{1-cosx}}{x}\end{array} \)
Solution:
\(\begin{array}{l}\lim _{x\rightarrow \:0}\frac{\sqrt{1-cosx}}{x}\\\mathrm{If\:}\lim _{x\to a-}f\left(x\right)=\lim _{x\to a+}f\left(x\right)=L\mathrm{\:then}\:\lim _{x\to a}f\left(x\right)=L\\\lim _{x\to \:0-}\left(\frac{\sqrt{1-\cos \left(x\right)}}{x}\right)\\=\lim _{x\to \:0-}\left(\frac{\sqrt{1-\cos \left(x\right)}}{x}\right)\\\lim _{x\to \:0-}\left(\frac{\sqrt{1-\cos \left(x\right)}}{x}\right)\\=\lim _{x\to \:0-}\left(\frac{\sqrt{\frac{\sin ^2\left(x\right)}{1+\cos \left(x\right)}}}{x}\right)\\=\lim _{x\to \:0-}\left(\frac{\frac{\left|\sin \left(x\right)\right|}{\sqrt{\cos \left(x\right)+1}}}{x}\right)\\=\lim _{x\to \:0-}\left(\frac{\frac{-\sin \left(x\right)}{\sqrt{\cos \left(x\right)+1}}}{x}\right)\\=\lim _{x\to \:0-}\left(-\frac{\sin \left(x\right)}{x\sqrt{\cos \left(x\right)+1}}\right)\\=-\lim _{x\to \:0-}\left(\frac{\sin \left(x\right)}{x\sqrt{\cos \left(x\right)+1}}\right)\\=-\lim _{x\to \:0-}\left(\frac{\cos \left(x\right)}{\sqrt{\cos \left(x\right)+1}-\frac{x\sin \left(x\right)}{2\sqrt{\cos \left(x\right)+1}}}\right)\\=-\lim _{x\to \:0-}\left(\frac{2\cos \left(x\right)\sqrt{\cos \left(x\right)+1}}{2\left(\cos \left(x\right)+1\right)-x\sin \left(x\right)}\right)\\=-\frac{2\cos \left(0\right)\sqrt{\cos \left(0\right)+1}}{2\left(\cos \left(0\right)+1\right)-0\cdot \sin \left(0\right)}\\=-\frac{1}{\sqrt{2}}\\\end{array} \)
Example 10: Evaluate \(\begin{array}{l}\lim_{x \rightarrow \infty} \frac{2e^{x}}{x}\end{array} \)
Solution:
Given \(\begin{array}{l}\lim_{x \rightarrow \infty} \frac{2e^{x}}{x}\end{array} \)
If we put x = ∞, the result is of the form ∞/∞.
Apply L’Hospital rule.
Differentiating the numerator and denominator, and we get,
\(\begin{array}{l}\lim_{x \rightarrow \infty} \frac{2e^{x}}{x}\end{array} \)
\(\begin{array}{l}=\lim_{x \rightarrow \infty} \frac{2e^{x}}{1}\end{array} \)
Now, we get \(\begin{array}{l}\lim_{x \rightarrow \infty} \frac{2e^{x}}{1}\end{array} \)
= ∞
Hence, the required solution is ∞.
Also read
Limits of Functions
Limits Continuity and Differentiability
Limits Solved Examples
L’Hospital’s Rule – Video Lesson
Frequently Asked Questions
Q1
When do you use L’Hospital’s rule?
When we have an indeterminate form 0/0 or ∞/∞ , we use L’ Hospital’s rule. We differentiate the numerator and differentiate the denominator and then determine the limit.
Q2
Give L’Hospital’s rule formula to evaluate indeterminate limits.
L’Hospital’s rule formula is given by
limx→ a f(x)/g(x) = limx→ a f’(x)/g’(x).
Q3
What do you mean by indeterminate forms?
The indeterminate form is a mathematical expression which means that we cannot determine the original value even after the substitution of the limits.
Q4
Give three indeterminate forms.
0/0, ∞/∞, 0×∞ are indeterminate forms.
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