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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

## 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.
\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

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!
\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

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

## Want to join the conversation?

• Why is 1^infinity an indeterminate form?
• This stems from the fact that all of the limits in calculus of this type have something to do with the number e (2.71828...)

e is actually defined as limit(n->infinity, (1+1/n)^n).
At first glance, (1+1/n) seems to be 1 and hence, this is called 1^infinity form. However, the limit of this quantity is 2.718...
• 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?
• 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)
• We multiply by π because (by the chain rule) we're multiplying by the derivative of xπ, the function inside of the sine function.
• what is the derivative of e^1/2 or e^0.5
(1 vote)
• e^0.5 is a constant, around 1.65. So the derivative is just 0.
(1 vote)
• 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)
• 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)
• 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)
• 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)
• 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!

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