Consider lim xln x) This is an indeterminate form of the
Consider lim x0+ (xlnx) This is an indeterminate form of the type 0 1 To apply l’H^opital’s rule we must rewrite it as a quotient First try: lim x0+ x (lnx)−1 is an indeterminate form of type 0
Limits involving ln(
Limits involving ln(x) We can use the rules of logarithms given above to derive the following information about limits lim x1 lnx = 1; lim x0 lnx = 1 : I We saw the last day that ln2 > 1=2 I Using the rules of logarithms, we see that ln2m = mln2 > m=2, for any integer m I Because lnx is an increasing function, we can make ln x as big as we
Indeterminate Forms - Department of Mathematics
x→∞ ln((1 + 1 x)x)) = exp( lim x→∞ xln(1 + 1 x)) = exp( lim x→∞ ln(1 + 1 x) 1/x) We can now apply L’Hopital’s since the limit is of the form 0 0 = exp( lim x→∞ (1/(1 + 1 x))(−1/x2) −1/x2) = exp( lim x→∞ 1/(1 + 1 x)) = exp(1) = e Exercises I Find the limits A lim x→∞ (1 + 1 x)3x B lim x→∞ (1 + k x)x C
Lecture 4 : Calculating Limits using Limit Laws
lim xaf(x) n, where nis a positive integer (we see this using rule 4 repeatedly) 7 lim xac= c, where c is a constant ( easy to prove from de nition of limit and easy to see from the graph, y= c) 8 lim xax= a, (follows easily from the de nition of limit) 9 lim xax n= an where nis a positive integer (this follows from rules 6 and 8) 10 lim
Limit as x Goes to Infinity of x(1/x) - MIT OpenCourseWare
x→∞ Solution This calculation is very similar to the calculation of lim x x presented in lecture, x→0+ except that 0instead of the indeterminate 0form 0 we instead have ∞ As before, we use the exponential and natural log functions to rephrase the problem: 1/x ln x 1 /x ln x x = e = e x Thus, lim x 1/x ln= lim e x x
Calculus Cheat Sheet - Lamar University
e & lim 0 x e 2 lim ln x x & 0 lim ln x x 3 If r 0 then lim 0 x r b x 4 If r 0 and xr is real for negative x then lim 0 x r b x 5 n even : lim n x x 6 n odd : lim n x x & lim n x x 7 n even : lim sgnn x ax bx c a 8 n odd : lim sgnn x ax bx c a 9 n odd : lim sgnn x ax cx d a
1 Definition and Properties of the Exp Function
e x= ex > 0 ⇒ E(x) = e is concave up, increasing, and positive Proof Since E(x) = ex is the inverse of L(x) = lnx, then with y = ex, d dx ex = E0(x) = 1 L0(y) = 1 (lny)0 = 1 1 y = y = ex First, for m = 1, it is true Next, assume that it is true for k, then d k+1 dxk+1 ex = d dx d dxk ex = d dx (ex) = ex By the axiom of induction, it is
1801 Single Variable Calculus Fall 2006 For information
Lecture 6 18 01 Fall 2006 M(a) (slope of ax at x=0) ax Figure 1: Geometric definition of M(a) 2 Geometrically, M( a) is theslope of graph y =x at x 0 The trick to figuring out what M(a) is is to beg the question and define e as the number such
Math 116 — Practice for Exam 2
Math 116 / Final (December 17, 2013) page 4 2 [11 points] Determine the convergence or divergence of the following series In parts (a) and (b), support your answers by stating and properly justifying any test(s), facts or computations
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