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

Question

1286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-26T10:47:27

#= 3/4#

#lim_{x \to 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

Wikipedia

so that 's

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

#= lim_{x \to 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#