May 2, 2016 · How do you find the limit of #(ln (ln (x) ) ) / ( ln (x) ) # as x approaches #1#? Calculus Limits Determining Limits Algebraically. 1 Answer

Jan 12, 2018 · But maybe the question is: lim_(x to oo) (x*ln((x+1)/(x-1))) that goes to oo*0 so we need to manipulate it ...

Apr 20, 2015 · Using the property of logarithm: log(ab)=loga+logb (with a,b>0) lim_(xrarr1)(ln(4x)-ln(x))/(ln(x))=lim_(xrarr1)(log4+logx-logx)/logx= =lim_(xrarr1)log4/logx=oo.

Dec 4, 2016 · Therefore, the use of L'Hôpital's rule is warranted: Compute the first derivative of the numerator: (d(x - 1 - ln(x)))/dx = 1 -1/x Compute the first derivative of the denominator: (d(ln(x)(x - 1)))/dx = (x - 1)/x + ln(x) Make a new fraction out of the new numerator and new denominator: lim_(xto1)[(1 -1/x)/((x - 1)/x + ln(x))] Multiply by x/x ...

It is quite simple. You must use the fact that ln(x) = e^(ln(ln(x))) Then, you know that ln(x)^(1/x) = e^(ln(ln(x))/x) And then, the interesting part happens which could be solved in two ways - using intuition and using maths. Let us start with intuition part. lim_(n->infty) e^(ln(ln(x))/x = lim_(n->infty)e^(("something smaller than x")/x) = e^0 = 1 Let us think why is that so? Thanks to ...

May 11, 2018 · What is the limit as x turns to infinity of #x(ln(x+3)-ln(x-3))#? Calculus. 1 Answer

Aug 15, 2016 · 1 lim_(x to oo) ln (x)/ln (3x) lim_(x to oo) ln (x)/(ln 3 + ln x ) lim_(x to oo) 1/(ln 3/ (ln x ) + 1 ) (lim_(x to oo) 1)/(lim_(x to oo) ln 3/ (ln x ) + 1 ) = 1/1 = 1 ...

Aug 2, 2016 · Use L'Hôpital's rule. Answer is: lim_(x->oo)ln(x+1)/x=0 lim_(x->oo)ln(x+1)/x This limit can't be defined ...

Dec 16, 2014 · lim_{x \to +oo} \frac{x^{ln 2}}{1+ln x} = +oo This can be found by using L'Hospital's Rule, which states that every limit of a fraction is equal to the limit of the derivatives of the fraction.

May 21, 2016 · We can rewrite the argument of the natural logarithm function: #lim_(xrarroo)ln(1+e^x)/(5x)=lim_(xrarroo)ln(e^x(e^-x+1))/(5x)#