Find the limit of f(x)?

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1 Answer
Steve M
Feb 2, 2018

# lim_(x rarr oo) f(x) = 3 #

Explanation:

Consider the limit:

# L_1 = lim_(x rarr oo) (6e^x-17)/(2e^x) #
# \ \ \ \ = lim_(x rarr oo) {(6e^x)/(2e^x)-17/(2e^x)} #
# \ \ \ \ = lim_(x rarr oo) {3-17/(2e^x)} #
# \ \ \ \ = 3-17/2 \ lim_(x rarr oo)e^-x #
# \ \ \ \ = 3-0 #
# \ \ \ \ = 3 #

Consider the limit:

# L_2 = lim_(x rarr oo) (3sqrt(x))/(sqrt(x-1)) #
# \ \ \ \ = 3 \ lim_(x rarr oo) sqrt(x/(x-1)) #
# \ \ \ \ = 3 \ lim_(x rarr oo) sqrt(x/(x-1) \ (1/x)/(1/x)) #
# \ \ \ \ = 3 \ lim_(x rarr oo) sqrt(1/(1-1/x) ) #
# \ \ \ \ = 3 \ sqrt(1/(1-0) ) #
# \ \ \ \ = 3 #

Now we are given that:

# (6e^x-17)/(2e^x) lt f(x) lt (3sqrt(x))/(sqrt(x-1)) #

And so we have:

# lim_(x rarr oo) (6e^x-17)/(2e^x) lt lim_(x rarr oo) f(x) lt lim_(x rarr oo) (3sqrt(x))/(sqrt(x-1)) #

Then using the above results we have:

# 3 lt lim_(x rarr oo) f(x) lt 3 #

And then by the Squeeze (or sandwich) Theorem, we deduce that:

# lim_(x rarr oo) f(x) = 3 #

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