What is the limit of #f(x) = (1 + 4e^-x )/( 3 - 2e^-x)# as x goes to infinity? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer Jim H Sep 26, 2015 #lim_(xrarroo)(1 + 4e^-x )/( 3 - 2e^-x) = 1/3# Explanation: As #xrarroo#, we have #e^-x rarr 0#, so #1+4e^-x rarr 1# while #3-2e^-x rarr 3#. Interestingly, as #xrarr-oo# we get a different limit. We have #lim_(xrarr-oo)(1 + 4e^-x )/( 3 - 2e^-x) = 4/-2 =-2 # Answer link Related questions What kind of functions have horizontal asymptotes? How do you find horizontal asymptotes for #f(x) = arctan(x)# ? How do you find the horizontal asymptote of a curve? How do you find the horizontal asymptote of the graph of #y=(-2x^6+5x+8)/(8x^6+6x+5)# ? How do you find the horizontal asymptote of the graph of #y=(-4x^6+6x+3)/(8x^6+9x+3)# ? How do you find the horizontal asymptote of the graph of y=3x^6-7x+10/8x^5+9x+10? How do you find the horizontal asymptote of the graph of #y=6x^2# ? How can i find horizontal asymptote? How do you find horizontal asymptotes using limits? What are all horizontal asymptotes of the graph #y=(5+2^x)/(1-2^x)# ? See all questions in Limits at Infinity and Horizontal Asymptotes Impact of this question 1659 views around the world You can reuse this answer Creative Commons License