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Unformatted text preview: 7x 2
x→∞
5x
The following uses the natural log function, L’Hopital’s Rule, and no ”tricks”.
Find the following limit: lim Let y = lim x→∞ 2
1+
5x 1+ 7x Now use the properties of the natural log function to write the exponential expression as a product.
Remember, the natural log function is continuous on (0, ∞). ln y = ln lim x→∞ 2
1+
5x 7x 7x 2
= lim ln 1 +
x→∞
5x We can write our product as a quotient by dividing by 2
ln y = lim 7x ln 1 +
x→∞
5x = lim 7x ln 1 +
x→∞ 2
5x 1
instead of multiplying by 7x .
7x
ln 1 + 2
5x = lim x→∞ 1
7x Now we have a quotient, so we can see if we can use L’Hopital’s Rule.
0
This is a ” ” form (remember, ln(1) = 0) so we can use L’Hopital’s Rule.
0
Note 1. Remember to use the chain rule for the derivative of ln 1 +
Note 2. For ease of diﬀerentiation, we can write
d2
=
dx 5x 2
5 ln 1 + So 2
5x lim x→∞ 1
7x −1
x2 and = lim x→∞ So now we have ln y = ln lim d1
=
dx 7x 1
1 + 52
x
1
7 x→∞ 2
1+
5x 2
=
5x 1
7 = 1
x and 1
=
7x 1
7 1
x −1
. Using L’Hopital’s Rule, we get:
x2 −1
2
5
x2
−1
x2
7x 2
5 2
.
5x 1
1 + 52
x
1
7 = lim x→∞ 14
.
5
1+ Finally, taking the exponential we get y = lim x→∞ 1 2
5x 7x
14 =e5. 2
5 2
5 1
= =
1
7 14
5 ...
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This note was uploaded on 09/25/2011 for the course CALCULUS 135 taught by Professor Augustarainsford during the Spring '11 term at Rutgers.
 Spring '11
 Augustarainsford

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