What is #\lim _ { x \rightarrow 0 } \frac { e ^ { 2 x } - e ^ { x } } { x }#?

1 Answer
Mar 3, 2017

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = 1 #

Explanation:

Method 1 : Graphically
graph{(e^(2x)-e^x)/(x) [-8.594, 9.18, -1.39, 7.494]}

Although far from conclusive, it appears that:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = 1#

Method 2 : L'Hôpital's rule

The limit:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) #

is of an indeterminate form #0/0#, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exits then:

# lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x)) #

And so applying L'Hôpital's rule we get:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = lim_(x rarr 0) (2e^(2x)-e^x)/(1)#
# " "= 2-1#
# " "= 1#

Method 3 - Power Series

The power series for #e^x# is as follows;

# e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... #

And so we have:

# e^(2x) = 1 + 2x + (2x)^2/(2!) + (2x)^3/(3!) + (2x)^4/(4!) + ... #

Therefore;

# (e^(2x)-e^x)/(x) = { (1 + 2x + O(x^2)) - (1 + x + O(x^2)) } / x #
# " " = { 1 + 2x + O(x^2) - 1 - x + O(x^2) } / x #
# " " = { x + O(x^2) } / x #
# " " = 1 + O(x) #

And so:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = lim_(x rarr 0) (1 + O(x)) #
# " " = 1 #