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.
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^(sin(x)) as x approaches 0 ... - Socratic
2016年9月27日 · #= 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 unnecessary, see ratnaker-m's note below... this is now in ...
What is the limit lim_(x->0)sin(x)/x? + Example - Socratic
2014年10月11日 · #lim_(x->0) sin(x)/x = 1#. We determine this by the use of L'Hospital's Rule. To paraphrase, L'Hospital's rule states that when given a limit of the form #lim_(x->a) f(x)/g(x)#, where #f(a)# and #g(a)# are values that cause the limit to be indeterminate (most often, if both are 0, or some form of #oo#), then as long as both functions are continuous and differentiable at and in the vicinity of ...
What is the limit of sin (1/x) as x approaches 0? - Socratic
2017年6月22日 · The limit does not exist. To understand why we can't find this limit, consider the following: We can make a new variable h so that h = 1/x. As x -> 0, h -> oo, since 1/0 is undefined. So, we can say that: lim_(x->0)sin(1/x) = lim_(h->oo)sin(h) As h gets bigger, sin(h) keeps fluctuating between -1 and 1. It never tends towards anything, or stops fluctuating at any point. So, we can say that the ...
What is the limit of sinx / x as x goes to infinity? - Socratic
2015年9月28日 · What happens to our function when #x# balloons up?. No matter what the input, #sinx# just oscillates between #0# and #1#.
How do you find the limit of sinx/(x-pi) as x->pi? - Socratic
2016年10月26日 · Use L'Hôpital's rule lim_(x→pi)sin(x)/(x - pi)= lim_(x→pi)cos(x) = -1 Use L'Hôpital's rule The derivative of the numerator is cos(x) The derivative of the denominator is 1 so we will not write it. lim_(x→pi)sin(x)/(x - pi) = lim_(x→pi)cos(x) = -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 ...
Special Limits Involving sin(x), x, and tan(x) - Calculus - Socratic
#lim_{x to 0}{sin x}/{x}=1#. (Note: This limit indicates that two functions #y=sin x# and #y=x# are very similar when #x# is near #0# .) Wataru · · Sep 7 2014
What is the limit of #(sin^2x/x^2)# as x approaches infinity?
2018年3月21日 · lim_(x->oo)sin^2x/x^2=0 For limits involving trigonometric functions divided by something, we're best off using the Squeeze Theorem, which tells us that if we have a function f(x), we can definite h(x) and g(x) such that h(x)<=f(x)<=g(x) And that if lim_(x->a)h(x)=lim_(x->a)g(x), then lim_(x->a)f(x)=lim_(x->a)g(x)=lim_(x->a)g(x). In other words, if the two functions between which f(x) lies ...
How do you evaluate the limit sin(5x)/x as x approaches 0
2016年11月7日 · How do you evaluate the limit #sin(5x)/x# as x approaches #0#? Calculus Limits Determining Limits Algebraically. 1 Answer