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L'Hôpital's rule review

L'Hôpital's rule helps us find many limits where direct substitution ends with the indeterminate forms 0/0 or ∞/∞. Review how (and when) it's applied.

What is L'Hôpital's rule?

L'Hôpital's rule helps us evaluate indeterminate limits of the form start fraction, 0, divided by, 0, end fraction or start fraction, infinity, divided by, infinity, end fraction.
In other words, it helps us find limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction, where limit, start subscript, x, \to, c, end subscript, u, left parenthesis, x, right parenthesis, equals, limit, start subscript, x, \to, c, end subscript, v, left parenthesis, x, right parenthesis, equals, 0 (or, alternatively, where both limits are plus minus, infinity).
The rule essentially says that if the limit limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction exists, then the two limits are equal:
limit, start subscript, x, \to, c, end subscript, start fraction, u, left parenthesis, x, right parenthesis, divided by, v, left parenthesis, x, right parenthesis, end fraction, equals, limit, start subscript, x, \to, c, end subscript, start fraction, u, prime, left parenthesis, x, right parenthesis, divided by, v, prime, left parenthesis, x, right parenthesis, end fraction
Want to learn more about L'Hôpital's rule? Check out this video.

Using L'Hôpital's rule to find limits of quotients

Let's find, for example, limit, start subscript, x, \to, 0, end subscript, start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, end fraction.
Substituting x, equals, 0 into start fraction, 7, x, minus, sine, left parenthesis, x, right parenthesis, divided by, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, end fraction results in the indeterminate form start fraction, 0, divided by, 0, end fraction. So let's use L’Hôpital’s rule.
=limx07xsin(x)x2+sin(3x)=limx0ddx[7xsin(x)]ddx[x2+sin(3x)]L’Hopital’s rule=limx07cos(x)2x+3cos(3x)=7cos(0)2(0)+3cos(30)Substitution=2\begin{aligned} &\phantom{=}\displaystyle\lim_{x\to 0}\dfrac{7x-\sin(x)}{x^2+\sin(3x)} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[7x-\sin(x)]}{\dfrac{d}{dx}[x^2+\sin(3x)]}\qquad\gray{\text{L'Hopital's rule}} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{7-\cos(x)}{2x+3\cos(3x)} \\\\ &=\dfrac{7-\cos(0)}{2(0)+3\cos(3\cdot0)}\qquad\gray{\text{Substitution}} \\\\ &=2 \end{aligned}
Note that we were only able to use L’Hôpital’s rule because the limit limit, start subscript, x, \to, 0, end subscript, start fraction, start fraction, d, divided by, d, x, end fraction, open bracket, 7, x, minus, sine, left parenthesis, x, right parenthesis, close bracket, divided by, start fraction, d, divided by, d, x, end fraction, open bracket, x, squared, plus, sine, left parenthesis, 3, x, right parenthesis, close bracket, end fraction actually exists.
Problem 1.1
limit, start subscript, x, \to, 0, end subscript, start fraction, e, start superscript, x, end superscript, minus, 1, divided by, 2, x, end fraction, equals, question mark
Choose 1 answer:
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Using L'Hôpital's rule to find limits of exponents

Let's find, for example, limit, start subscript, x, \to, 0, end subscript, left parenthesis, 1, plus, 2, x, right parenthesis, start superscript, start superscript, start fraction, 1, divided by, sine, left parenthesis, x, right parenthesis, end fraction, end superscript, end superscript. Substituting x, equals, 0 into the expression results in the indeterminate form 1, start superscript, start superscript, infinity, end superscript, end superscript.
To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting y, equals, left parenthesis, 1, plus, 2, x, right parenthesis, start superscript, start superscript, start fraction, 1, divided by, sine, left parenthesis, x, right parenthesis, end fraction, end superscript, end superscript, we will find limit, start subscript, x, \to, 0, end subscript, natural log, left parenthesis, y, right parenthesis. Once we find it, we will be able to find limit, start subscript, x, \to, 0, end subscript, y.
natural log, left parenthesis, y, right parenthesis, equals, start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction
Substituting x, equals, 0 into start fraction, natural log, left parenthesis, 1, plus, 2, x, right parenthesis, divided by, sine, left parenthesis, x, right parenthesis, end fraction results in the indeterminate form start fraction, 0, divided by, 0, end fraction, so now it's L’Hôpital’s rule's turn to help us with our quest!
=limx0ln(y)=limx0ln(1+2x)sin(x)=limx0ddx[ln(1+2x)]ddx[sin(x)]L’Hopital’s rule=limx0(21+2x)cos(x)=(21)1Substitution=2\begin{aligned} &\phantom{=}\displaystyle\lim_{x\to 0}\ln(y) \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\ln(1+2x)}{\sin(x)} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\ln(1+2x)]}{\dfrac{d}{dx}[\sin(x)]}\qquad\gray{\text{L'Hopital's rule}} \\\\ &=\displaystyle\lim_{x\to 0}\dfrac{\left(\dfrac{2}{1+2x}\right)}{\cos(x)} \\\\ &=\dfrac{\left(\dfrac{2}{1}\right)}{1}\qquad\gray{\text{Substitution}} \\\\ &=2 \end{aligned}
We found that limit, start subscript, x, \to, 0, end subscript, natural log, left parenthesis, y, right parenthesis, equals, 2, which means limit, start subscript, x, \to, 0, end subscript, y, equals, e, squared.
Problem 2.1
limit, start subscript, x, \to, 0, end subscript, open bracket, cosine, left parenthesis, 2, pi, x, right parenthesis, close bracket, start superscript, start superscript, start fraction, 1, divided by, x, end fraction, end superscript, end superscript, equals, question mark
Choose 1 answer:
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • starky ultimate style avatar for user ∆
    Why is 1^infinity an indeterminate form?
    (3 votes)
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  • male robot hal style avatar for user Yash
    In the article's example for using L'Hopital's rule for finding limits of exponents, they get (1+2(0))^1/sin(0) = 1^infinity (direct substitution). But won't 1/sin(0) be undefined, thus resulting in 1^undefined = undefined?
    (3 votes)
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  • duskpin ultimate style avatar for user khan academy
    In the explanation to problem 1.2, the derivative of the top equation is shown as

    d/dx​[xcos(πx)] = cos(xπ) - πxsin(xπ)
    .

    Where does the π in - πxsin... come from?
    (1 vote)
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  • duskpin tree style avatar for user Fahimuzzaman
    what is the derivative of e^1/2 or e^0.5
    (1 vote)
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  • piceratops ultimate style avatar for user hummusw
    Is there a printable version of this page?
    (1 vote)
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    • duskpin ultimate style avatar for user ondraperny
      Unfortunately there is nothing like that. So far best solution might be using Snipping tool on windows which can easily cut "pictures" from browser and then you can arrange them together in some software( Microsoft word would suffice).
      (1 vote)
  • blobby green style avatar for user Osmis
    I had a problem (1-4/x)^x . My question was when they took (4/x^2)/((1-4/x)(-1/x^2)) and got (4x^2)/(1-4/x)(-x^-2).
    (1 vote)
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  • male robot hal style avatar for user Vincent Pace
    When using L'Hôpital's rule to find limits of exponents, there's a step that sets, for example, lim x->∞ ln(y) equal to ln (lim x->∞ y). Which logarithm or limit property allows this?
    (1 vote)
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    • leafers ultimate style avatar for user Paras Sharma
      Here we can use this property because here we are not applying the limit to whole ln(y(x)) operator we have our variable x in the y(x) , So here we just wanna find the limiting value of y(x)
      It doesn't violate our previous method that we use we just plug the value and try to come up w/ a more subtle and concrete way of understanding this.
      (1 vote)
  • aqualine ultimate style avatar for user John He
    What about lim x→0 cot(x)/In(x)?If you apply L'Hôpital's rule,try to differentiate this,you will get into great trouble!

    ​​
    (0 votes)
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    • leaf green style avatar for user kubleeka
      Direct substitution gives ∞/∞, so taking the derivatives of according to l'Hôpital yields
      -csc²(x)/(1/x). This rearranges into -x/sin²(x).
      Direct substitution now yields 0/0, so we can apply l'Hôpital's Rule again. Differentiate to get -1/(2sin(x)cos(x))

      Now, finally, direct substitution yields -1/0, which indicates that the limit does not exist.
      (2 votes)