What is the limit of # sinx /(x^2 - 4x)# as x approaches 0?
2 Answers
Explanation:
You know that
#color(red)(cancel(color(black)(lim_(x->0)(sin0)/(0^2 - 4 * 0) = 0/0))) -># indeterminate form
This means that you're going to have to use L'Hopital's Rule to find this limit. According to L'Hopital's rule, if you have two functions
#lim_(x->c)(f^'(x))/(g^'(x))" "# , with#g^'(x)!=0# and#c in (a,b)#
exists, then you have
#color(blue)(lim_(x->0)(f(x))/(g(x)) = lim_(x->0)(f^'(x))/(g^'(x))#
In your case, you have
#d/dx(sinx) = cosx" "# and#" "d/dx(x^2-4x) = 2x-4#
which means that
#lim_(x->0)sinx/(x^2-4x) = lim_(x->0)(d/dx(sinx))/(d/dx(x^2-4x)) = cosx/(2x-4)#
Now you can evaluate this limit for
#lim_(x->0)(cos0)/(2 * 0 - 4) = 1/((-4)) = -1/4#
Therefore
#lim_(x->0)sinx/(x^2-4x) = color(green)(-1/4)#
Explanation:
# = lim_(xrarr0) sinx/x 1/((x-4))#
# = lim_(xrarr0) sinx/x lim_(xrarr0) 1/((x-4))# #" "# if both limits exist
# = (1)(-1/4)#
# = -1/4#