How do you find the limit of sin(x-1)/(x^2+x-2) as x->1? - Socratic
Oct 6, 2016 · 1/3 sin(x-1)/(x^2+x-2)= sin(x-1)/((x+2)(x-1)) = (sin(x-1)/(x-1)) 1/(x+2) lim_(x->1)sin(x-1)/(x^2+x-2)=lim_(x->1)(sin(x-1)/(x-1)) 1/(x+2) = 1 cdot 1/3
How do you use L'hospital's rule to find the limit lim_(x->0)(x-sin(x ...
Apr 11, 2018 · The limit. #L = lim_(x->0)(x-sin(x))/(x-tan(x))# is of the form #0/0#, Thus, we can use L'hospital's rule, which says
What is #lim x->0# of #((sqrt(1+sinx) -1) / (x))#? - Socratic
Oct 6, 2016 · #(sqrt(1+sinx) -1) / (x)= ((sqrt(1+sinx) -1)(sqrt(1+sinx) +1)) /( x(sqrt(1+sinx) -1)) = (1+sinx-1)/( x(sqrt(1+sinx) +1)) = sin x/x 1/(sqrt(1+sinx) +1)#
How do you use the Squeeze Theorem to find lim ( sin (2pi/x)) as x ...
Oct 12, 2015 · lim_(xrarr0) sin((2pi)/x) Does Not Exist Here is the key idea. However close to 0 we start, it will happen that as x rarr0, we get infinitely many values of (2pi)/x that are coterminal with pi/2, so they have sin((2pi)/x)=1. (Specifically, for every x=4/(1+4n with n an integer.) We also get infinitely many values of (2pi)/x that are coterminal with (3pi)/2, so they have sin((2pi)/x)=-1 ...
How do you evaluate the limit sin (6x)/cos (4x) as x ... - Socratic
Sep 29, 2017 · lim_(x rarr 0) sin(6x)/cos(4x) = 0 We seek: L = lim_(x rarr 0) sin(6x)/cos(4x) We note that both sintheta and cos theta are both continuous well behaved function and that both are defined when theta =0 Thus: L = (lim_(x rarr 0) sin(6x))/(lim_(x rarr 0) cos(4x)) = sin0/cos0 = 0/1 = 0
How do you evaluate the limit x/sin(x/2) as x approaches 0
Sep 10, 2016 · 2 Our destination is well-known limit usually written as lim_(z to 0) (sin z)/z = 1 but we can flip it: lim_(z to 0) z/(sin z) = 1 quad triangle So lim_(x to 0) x/sin(x/2) = lim_(x to 0) 2 (x/2)/sin(x/2) With sub y = x/2 and lifting the constant outside the limit =2 lim_(y to 0) y/sin y Now see triangle =2 * 1 = 2
How do you find the limit of #(sin(x)/3x) - Socratic
Mar 2, 2016 · lim_(x->0)(sin(x)/(3x)) = 1/3. See explanation below. lim_(x->0)(sin(x)/(3x)) = "0/0" so we can use Bernouilli L'Hôpital's rule. lim_(x->0)(sin(x)/(3x)) = lim_(x->0 ...
How do you find the limit of #sin(x+4sinx)# as x approaches pi?
May 18, 2015 · lim_(x->pi) sin(x + 4sinx) = sin(pi + 4sinpi) we know that color(red)(sinpi = 0) thus we have: = sin(pi + 4(0)) = sinpi =0
What is the limit of (sinx)(ln4x) as x approaches 0? - Socratic
Sep 9, 2016 · 0 when written as lim_(x to 0) (ln 4x)/(1/(sin x)) this is oo/oo indeterminate so we can use L'Hopital's Rule = lim_(x to 0) (1/x)/(- cscx cot x) = - lim_(x to 0) sin x tan x * (1/x) using well known limit: lim_(z to 0) (sin z)/z = 1 = - lim_(x to 0) (sin x)/x lim_(x to 0) tan x = - 1 * 0 = 0
How do you use continuity to evaluate the limit sin(x+sinx)?
Sep 26, 2015 · See explanation Long version: Sine is continuous so lim_(xrarra)sinx = sina x is continuous so, lim_(xrarra)x = a The sum of continuous functions is continuous, so lim_(xrarra)(x+sinx) = lim_(xrarra)x+ lim_(xrarra)sinx = a+sina The composition of continuous functions is continuous, so lim_(xrarra)(sin(x+sinx)) = sin(lim_(xrarra) (x+sinx)) = sin(a+sina) Short version: Because of various facts ...