What is the limit as x approaches infinity of #sin(x)#? - Socratic
2014年8月14日 · As x approaches infinity, the y-value oscillates between 1 and -1; so this limit does not exist. Thus, the answer is it DNE (does not exist). One good rule to have while solving these problems is that generally, if there is no x in the denominator at all, then the limit does not exist. Example: lim_(x->oo)sinx=DNE lim_(x->oo)(sinx)/(x)=0 (Squeeze Theorum) This is the same question as below ...
How do you find the limit of x/sinx as x approaches 0? - Socratic
2016年3月3日 · #lim_{x rarr 0} x/{sin x} = lim_{x rarr 0} 1/{cos x} = 1/{cos 0} = 1/1 = 1#. NOTE The question was posted in " Determining Limits Algebraically " , so the use of L'Hôpital's rule is NOT a suitable method to solve the problem.
How do you find the limit of x^(sin(x)) as x approaches 0 ... - Socratic
2016年9月27日 · 1 let L = lim_(x to 0) x^(sin x) implies ln L = ln lim_(x to 0) x^(sin x) = lim_(x to 0) ln x^(sin x) = lim_(x to 0) sinx ln x = lim_(x to 0) (ln x)/(1/(sinx) ) = lim_(x to 0) (ln x)/(csc x ) this is in indeterminate oo/oo form so we can use L'Hôpital's Rule = lim_(x to 0) (1/x)/(- csc x cot x) =- lim_(x to 0) (sin x tan x)/(x) Next bit is ...
What is the limit as x approaches infinity of sinx? | Socratic
2015年6月6日 · The range of y = sinx is R = [-1;+1]; the function oscillates between -1 and +1. Therefore, the limit when x approaches infinity is undefined.
How do you use the Squeeze Theorem to find lim Sin (x)/x as x ...
2015年10月19日 · #cosx < sinx/x < 1# for #-pi/2 < x < 0#. #lim_(xrarr0^-) cosx = 1# and #lim_(xrarr0^-) 1= 1# so #lim_(xrarr0^-) sinx/x = 1# Since both one sided limits are #1#, the limit is #1#. Note. This proof uses the fact that #lim_(xrarr0)cosx = 1#. That can also be stated "the cosine function is continuous at #0# ". That fact can be proved from the fact ...
What is the limit of sinx / x as x goes to infinity? - Socratic
2015年9月28日 · No matter what the input, #sinx# just oscillates between #0# and #1#. As the denominator gets larger and larger, we will be dividing by a larger number, which yields a smaller number. Since the numerator stays relatively the same, and the denominator blows up, #sinx/x# will become infinitesimally small and approach zero. Hope this helps!
How do you find lim sinx/x as x->0 using l'Hospital's Rule? - Socratic
2018年3月15日 · #lim_(x->0) sinx/x# is a well known standard limit #=1#. Here, we are asked to use l'Hospital's rule since the limit reduces to the indeterminate form #0/0# Thus, #lim_(x->0) sinx/x = lim_(x->0) (d/dx sinx)/(d/dx x)# #= lim_(x->0) cosx/1 = 1/1 =1#
Evaluate the limit? : lim_(x rarr 0) ( tanx-x ) / (x-sinx) - Socratic
2017年5月7日 · lim_(x rarr 0) ( tanx-x ) / (x-sinx) = 2 We want to find: L = lim_(x rarr 0) ( tanx-x ) / (x-sinx) Method 1 : Graphically graph{( tanx-x ) / (x-sinx) [-8.594, 9.18, -1.39, 7.494]} Although far from conclusive, it appears that: L = 2 Method 2 : L'Hôpital's rule The limit is of an indeterminate form 0/0, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then ...
Lim x->0 (tan x- sin x)/x^3=? - Socratic
2017年11月9日 · lim_(x->0) (tanx-sinx)/x^3 = 1/2 Transform the function in this way: (tanx-sinx)/x^3 = 1/x^3(sinx/cosx-sinx) (tanx-sinx)/x^3 = 1/x^3((sinx-sinxcosx)/cosx) (tanx-sinx ...
How do you use the Squeeze Theorem to find lim (1-cos(x))/x as x ...
2015年9月24日 · The usual procedure is to use the squeeze theorem (and some geometry/trigonometry) to prove that lim_(xrarr0)sinx/x=1 Then use that result together with (1-cosx)/x = sin^2x/x(1+cosx) = sinx/x sinx/(1+cosx) along with continuity of sine and cosine at 0 to get lim_(xrarr0)sinx/x sinx/(1+cosx) = 1 * 0/2 =0. So we can use the same geometric arguments to get the same bounds on sinx/x for small ...