2017年11月5日 · How do you find the limit #lnx/x# as #x->oo#? Calculus Limits Determining Limits Algebraically. 2 Answers

2015年3月24日 · You know that if #x>1 ln(x)>0# so the limit must be positive. You also know that #ln(x_2)-ln(x_1)=ln(x_2/x_1)# so if #x_2>x_1# the difference is positive, so #ln(x)# is always …

2016年8月27日 · 0 You can tell just by inspection that the limit will be zero for the simple reason that log(x) grows more slowly than x, and here it is actually log(log(x)) in the numerator. You …

2016年5月2日 · How do you find the limit of #(ln (ln (x) ) ) / ( ln (x) ) # as x approaches #1#? Calculus Limits Determining Limits Algebraically. 1 Answer

2016年8月1日 · lim_(xrarroo) (ln(x))^(1/x) = 1 We start with quite a common trick when dealing with variable exponents. We can take the natural log of something and then raise it as the …

2017年6月29日 · There is no limit as x approaches 0 from below since ln x is undefined for negative numbers. Instead, I will demonstrate how to find the right-handed limit, i.e., as x …

2018年1月3日 · lim_(x->1)ln(x)/(x-1)=1 First, we can try directly pluggin in x: ln(1)/(1-1)=0/0 However, the result 0 \\/ 0 is inconclusive, so we need to use another method. In this case, my …

2016年7月22日 · lim_(x->∞) ln(x+1)/(x) = 0 This limit is indeterminate because direct substitution yields ∞/∞. Therefore, we can apply L'Hospital's rule, which basically is taking a derivative of …

2016年6月21日 · #x - ln(x)# #=ln (e^x) + ln(x^{-1})# #= ln(e^x/x)# Already it should be clear that this is going to #infty# as the exponential is of greater order. To be clear we are now actually …

2017年2月6日 · lim_(x->0) lnx = -oo First we prove that ln(x) is monotone increasing. Consider: x_1, x_2 in RR^+ with x_2 > x_1 ln x_2 = ln (x_2/x_1*x_1) = ln( x_2/x_1) +ln x_1 x_2 > x_1 => …