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.
Log in ∆ 4 years agoPosted 4 years ago. Direct link to ∆'s post “Why is 1^infinity an inde...” Why is 1^infinity an indeterminate form? • (4 votes) The #1 Pokemon Proponent 4 years agoPosted 4 years ago. Direct link to The #1 Pokemon Proponent's post “This stems from the fact ...” 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). (5 votes) Yash 6 years agoPosted 6 years ago. Direct link to Yash's post “In the article's example ...” 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? • (4 votes) khan academy 4 years agoPosted 4 years ago. Direct link to khan academy's post “In the explanation to pro...” In the explanation to problem 1.2, the derivative of the top equation is shown as Where does the • (1 vote) kubleeka 4 years agoPosted 4 years ago. Direct link to kubleeka's post “We multiply by π because ...” We multiply by π because (by the chain rule) we're multiplying by the derivative of xπ, the function inside of the sine function. (4 votes) Fahimuzzaman 2 years agoPosted 2 years ago. Direct link to Fahimuzzaman's post “what is the derivative of...” what is the derivative of e^1/2 or e^0.5 • (1 vote) kubleeka 2 years agoPosted 2 years ago. Direct link to kubleeka's post “e^0.5 is a constant, arou...” e^0.5 is a constant, around 1.65. So the derivative is just 0. (2 votes) 470050 8 months agoPosted 8 months ago. Direct link to 470050's post “i watched the video.” i watched the video. • (1 vote) adt 22 days agoPosted 22 days ago. Direct link to adt's post “in the exercises, I was c...” in the exercises, I was confused how: ln(y) =(x-1)ln(x-1) could someone help? • (1 vote) Aeges97 12 days agoPosted 12 days ago. Direct link to Aeges97's post “(x-1) is the same as 1/(x...” (x-1) is the same as 1/(x-1)^-1. Putting it in that form makes it useful for checking L'Hopital's Rule because we don't care about the overall function/quotient to start with, just the individual functions themselves. The limit of (x-1)^-1 = 1/(x-1) as x approaches 1 from the positive direction is infinity, which is readily apparent if you graph it on Desmos or some such. Going through L'Hopital's rule you'll eventually get the limit of Ln(y) = 0, so for that to be true the limit of y, and thus the limit we want, must be 1. Hope that helps. (1 vote) hummusw 8 years agoPosted 8 years ago. Direct link to hummusw's post “Is there a printable vers...” Is there a printable version of this page? • (1 vote) ondraperny 7 years agoPosted 7 years ago. Direct link to ondraperny's post “Unfortunately there is no...” 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) Osmis 2 years agoPosted 2 years ago. Direct link to Osmis's post “I had a problem (1-4/x)^x...” 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) Vincent Pace 7 years agoPosted 7 years ago. Direct link to Vincent Pace's post “When using L'Hôpital's ru...” 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) Paras Sharma 7 years agoPosted 7 years ago. Direct link to Paras Sharma's post “Here we can use this prop...” 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) (1 vote) John He 7 years agoPosted 7 years ago. Direct link to John He's post “What about lim x→0 cot(x)...” 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) kubleeka 7 years agoPosted 7 years ago. Direct link to kubleeka's post “Direct substitution gives...” Direct substitution gives ∞/∞, so taking the derivatives of according to l'Hôpital yields Now, finally, direct substitution yields -1/0, which indicates that the limit does not exist. (2 votes)Want to join the conversation?
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...
.d/dx[xcos(πx)] = cos(xπ) - πxsin(xπ)π in - πxsin... come from?
became: ln(y)=( ln(x-1) ) / (x-1)^-1
the question was to find the limit as x approaches 1 from the positive side for (x-1)^(x-1)
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.
-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))
