How do you find #lim sin(2x)/x# as #x->0# using l'Hospital's Rule? Calculus Tests of Convergence / Divergence Indeterminate Forms and de L'hospital's Rule 1 Answer Andrea S. Mar 1, 2017 #lim_(x->0) sin(2x)/x = 2# Explanation: The limit: #lim_(x->0) sin(2x)/x# is in the indeterminate form #0/0# so we can solve it using l'Hospital's rule: #lim_(x->0) sin(2x)/x = lim_(x->0) (d/dx sin(2x))/(d/dx x) = lim_(x->0) (2cos(2x))/1 = 2# Answer link Related questions What is L'hospital's Rule? How do you use L'hospital's rule to find the limit? What if L'hospital's rule doesn't work? What is L'hospital's rule used for? How do you know when to not use L'hospital's rule? How do you know when to use L'hospital's rule twice? How do you use L'hospital's rule to find the limit #lim_(x->oo)ln(x)/sqrt(x)# ? How do you use L'hospital's rule to find the limit #lim_(x->oo)x^3e^(-x^2)# ? How do you use L'hospital's rule to find the limit #lim_(x->oo)xsin(pi/x)# ? How do you use L'hospital's rule to find the limit #lim_(x->0)(x-sin(x))/(x-tan(x))# ? See all questions in Indeterminate Forms and de L'hospital's Rule Impact of this question 38209 views around the world You can reuse this answer Creative Commons License