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#

2014年9月21日 · The answer is oo. The natural log function is strictly increasing, therefore it is always growing albeit slowly. The derivative is y'=1/x so it is never 0 and always positive. You can also look at it as: n=ln oo e^n=oo Therefore, n must be large.

2016年6月21日 · combine terms and use L'Hopital's Rule is one way 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 ...

2016年5月21日 · We can also analyze this intuitively: the linear function #n# rises at a greater rate than the logarithmic function #ln(n)#, so since the function that rises faster is in the denominator, the function will approach #0#. If the function had been flipped, we'd see that the limit as #n# approaches #oo# in #n/ln(n)# would be #oo#.

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

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

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 ...

2015年2月16日 · First of all it is important to say that oo, without any sign in front of, would be interpreted as both, and it is a mistake! The argument of a logarithmic function has to be positive, so the domain of the function y=lnx is (0,+oo). So: lim_(xrarr+oo)lnx=+oo, as shown by the graphic. graph{lnx [-10, 10, -5, 5]}

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8月6日 · Make the limit of (1+(1/x))^x as x approaches infinity equal to any variable e.g. y, k. and take the natural logarithm of both sides. y=lim_(x-oo)(1+(1/x))^x ln y ...