L'Hospital's Rule for Solving Indeterminate Limits

In calculus, L’ Hospital’s rule is a powerful tool to evaluate limits of indeterminate forms. This rule will be able to show that a limit exists or not, if yes then we can determine its exact value. In short, this Rule tells us that in case we are having indeterminate forms like 0/0 and ∞/∞ then we just differentiate the numerator as well as the denominator and simplify evaluation of limits. This article helps students to learn how to use L Hospitals rule for solving indeterminate limits.

What do you mean by Indeterminate Forms- 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, ie 0/0 and ∞/∞.

Solved Problems

Example 1: Find limx3x2x6x3\lim_{x\to 3} \frac{x^{2}-x-6}{x-3}

Solution: 

Given  limx3x2x6x3\lim_{x\to 3} \frac{x^{2}-x-6}{x-3}

If we put x = 3 , we get a solution of the form 0/0.

Apply L’hospital rule.

 Differentiate numerator and denominator we get, 

limx3x2x6x3\lim_{x\to 3} \frac{x^{2}-x-6}{x-3} = limx32x11\lim_{x\to 3} \frac{2x-1}{1}

= [2(3)-1]/1 = (6-1)/1 [substitute x = 3]

= 5

Hence the required solution is 5.

Example 2: Solve limx0sin xxlim_{x \rightarrow 0} \frac{sin\ x }{x}

Solution:

Put x = 0 in the given function we get 0/0.

To solve this question we just differentiate numerator and denominator separately.

limx0cos x1lim_{x \rightarrow 0} \frac{cos\ x }{1} = cos(0)/1 = 1/1 = 1

Example 3: Solve limx0excos xsin 2xlim_{x \rightarrow 0} \frac{e^x – cos\ x }{sin\ 2x}

Solution:

At x = 0, given function is in 0/0 form.

On differentiating given function, we get

limx0exsin x2 cos 2xlim_{x \rightarrow 0} \frac{e^x – sin\ x }{2\ cos\ 2x} = 1/2

Example 4: Find limit for below functions:

(i))limx4x25x3x21(ii)limx15x44x219x3+x10(iii)limx0+e3x5x+200(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}\\

Solution:

(i) At x = ∞, given function is in ∞/∞ form.

On differentiating given function twice, we have

limx4x25x3x21lim_{x \rightarrow \infty} \frac{4x^2-5x}{3x^2-1} = limxlim_{x \rightarrow \infty} 8/6 = 4/3

(ii) At x = 1, given function is in 0/0 form.

Differentiate given function,

limx120x38x27x2+1lim_{x \rightarrow 1} \frac{20x^3-8x}{27x^2+1} = 12/28 = 3/7

(iii) limx0+3e3x5lim_{x \rightarrow 0^+} \frac{3e^{3x}}{5} = 3/5

Example 5: Evaluate limx0+x1/3ln(x)\lim_{x \rightarrow 0^+} \frac{x^{-1/3}}{ln(x)}

Solution:

limx0+x1/3ln(x)\lim_{x \rightarrow 0^+} \frac{x^{-1/3}}{ln(x)}

L Hospital Rule Examples

Example 6: Evaluate limit for the function, limx0+sin2x5x\lim_{x\rightarrow 0^+}\frac{sin2x}{\sqrt{5x}}

Solution:

Examples on L Hospital Rules

Example 7: Solve limit for the function applying L’Hospital’s Rule.

limx01costt3\lim_{x\rightarrow 0^{-}}\frac{1-cost}{t^{3}}

Solution:

limx01costt3=1cos00=00\lim_{x\rightarrow 0^{-}}\frac{1-cost}{t^{3}} = \frac{1-cos0}{0}=\frac{0}{0} (Indeterminate form)

Applying L’Hospital’s Rule, we have

limx01costt3=limt0(sin(t)3t2)=limt0(cos(t)6t)=16limt0(cos(t)t)=16limt0(cos(t)1t)=16limt0(cos(t))limt0(1t)limt0(cos(t))=cos(0)=1limt0(1t)==161()=\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 \:\\

Example 8: Evaluate the limit limx1(1x+ln(x)1+cos(πx))\lim _{x\to \:1-}\left(\frac{1-x+\ln \left(x\right)}{1+\cos \left(\pi x\right)}\right)

Solution: 

limx1(1x+ln(x)1+cos(πx))\lim _{x\to \:1-}\left(\frac{1-x+\ln \left(x\right)}{1+\cos \left(\pi x\right)}\right)

Apply L’Hospital’s Rule,

limx1(1x+ln(x)1+cos(πx))=limx1(1x1πsin(πx))=1x1πsin(πx)=x+1xπsin(πx)=x+1πxsin(πx)=limx1(1xπxsin(πx))=limx1(1π(sin(πx)+πxcos(πx)))=1π(sin(π1)+π1cos(π1))π(sin(π1)+π1cos(π1))=π2=1π2=1π2\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}\\

Example 9: Evaluate limx01cosxxdx\lim _{x\rightarrow \:0}\frac{\sqrt{1-cosx}}{x}dx

Solution:

limx01cosxxdxIflimxaf(x)=limxa+f(x)=Lthenlimxaf(x)=Llimx0(1cos(x)xdx)=limx0(1cos(x)x)limx0(dx)limx0(1cos(x)x)=limx0(sin2(x)1+cos(x)x)=limx0(sin(x)cos(x)+1x)=limx0(sin(x)cos(x)+1x)=limx0(sin(x)xcos(x)+1)=limx0(sin(x)xcos(x)+1)=limx0(cos(x)cos(x)+1xsin(x)2cos(x)+1)=limx0(2cos(x)cos(x)+12(cos(x)+1)xsin(x))=2cos(0)cos(0)+12(cos(0)+1)0sin(0)=12=(12)d0=0limx0+(1cos(x)xdx)=0\lim _{x\rightarrow \:0}\frac{\sqrt{1-cosx}}{x}dx\\\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}dx\right)\\=\lim _{x\to \:0-}\left(\frac{\sqrt{1-\cos \left(x\right)}}{x}\right)\cdot \lim _{x\to \:0-}\left(dx\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}}\\=\left(-\frac{1}{\sqrt{2}}\right)d0\\=0\\\lim _{x\to \:0+}\left(\frac{\sqrt{1-\cos \left(x\right)}}{x}dx\right)=0\\

Example 10: Evaluate limx2exx\lim_{x \rightarrow \infty} \frac{2e^{x}}{x}

Solution:

Given limx2exx\lim_{x \rightarrow \infty} \frac{2e^{x}}{x}

If we put x = ∞, the result is of the form ∞/∞.

Apply L’hospital rule.

Differentiating numerator and denominator we get, 

 limx2exx\lim_{x \rightarrow \infty} \frac{2e^{x}}{x} = limx2ex1\lim_{x \rightarrow \infty} \frac{2e^{x}}{1}

Now we get  limx2ex1\lim_{x \rightarrow \infty} \frac{2e^{x}}{1} = ∞ 

Hence the required solution is ∞.

Also read

Limits of functions

Limits Continuity and Differentiability

Limits Solved examples

 

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