How do you evaluate #(3x^2+2x+sinx)/(sinx+e^3x-1)# as x approaches 0?

Question

Emilia Aguilar · Accepted Answer

#= 3/4#

#lim_{x o 0} \ (3x^2+2x+sinx)/(sinx+e^color{red}{(3x)}-1)#

NB note the red correction, is there a typo in the question cos it looks odd to me?!?!

start by plugging on x = 0 to see what it looks like

that's

#(0+0+0)/(0+1-1) = 0/0# indeterminate so L'Hopital might be the way

so that 's

#lim_{x o 0} \ (3x^2+2x+sinx)/(sinx+e^color{red}{(3x)}-1)#

#= lim_{x o 0} \ (6x+2+cosx)/(cosx+3e^(3x))# by L'Hopital

Again we can pop in x = 0 to see where we are

that's

# (0+2+1)/(1+3) = 3/4#

Scarlett Allison · Answer

undefined To evaluate the expression (3x^2+2x+sinx)/(sinx+e^3x-1) as x approaches 0, we can substitute 0 for x in the expression. This gives us (3(0)^2+2(0)+sin(0))/(sin(0)+e^3(0)-1). Simplifying further, we have (0+0+0)/(0+1-1). Continuing to simplify, we get 0/0. However, this is an indeterminate form. To evaluate it further, we can use L'Hôpital's rule. Taking the derivative of the numerator and denominator separately, we get (6x+2+cosx)/(cosx+3e^3x). Substituting x=0 into this derivative expression, we have (6(0)+2+cos(0))/(cos(0)+3e^3(0)). Simplifying, we get (2+1)/(1+3). Therefore, the limit of the expression as x approaches 0 is 3/4.