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 growing. If #lim_{x->infty}ln(x) = M in RR# you have #ln(x)< M => x < e^M#, but #x->infty# so #M# can not be in #RR#, and the limit must be #+infty#

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 can also nail it more formally with L'Hopital's Rules, as it is oo/oo indeterminate lim_(x to oo) ln(ln(x))/x = lim_(x to oo) ( 1/ ln(x)* 1/x )/1 = lim_(x to oo) 1/ ln(x) * lim_(x to oo) 1/x = 0

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 exponent of the exponential function without changing its value as these are inverse operations - but it allows us to use the rules of logs in a beneficial way. lim_(xrarroo) (ln(x))^(1/x) = lim_(xrarroo) exp(ln((ln(x))^(1/x ...

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->0^+. Here is a graph: So we should expect the answer to be zero. Now, to do this we can't use the product rule, since the limit of ln x diverges as x->0^+, we have to be more clever. L'HOPITAL'S RULE: You can Google the precise ...

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 method of choice would be L'Hôpital's rule. It says that you if you have a limit resulting in the indeterminate form 0/0, you can differentiate both the numerator and the denominator, and if the limit of this exists ...

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 the numerator and the denominator at the same time. lim_(x->∞) ln(x+1)/(x) -> ∞/∞ Applying L'Hospital's rule gives us lim_(x->∞) 1/(x+1) = 0 This makes …

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 looking inside the log at #z = lim_{x \to \infty} e^x /x # and the cheapest shot is using L'Hopital's rule as this #\infty / \infty# form is indeterminate. By L ...

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 => x_2/x_1 > 1 => ln(x_2/x_1) > 0 => ln(x_2) > ln(x_1) which proves the point. Since it is monotone increasing lnx has a limit for x->oo and since the function is not bounded this limit must be +oo, so: lim_(x->oo) lnx = +oo ...