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L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
Theorem: Let k > 0, then lim k1/x = 1.x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K K1/2
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K K1/2 K1/3
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1K K1/2 K1/3 K1/4
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK K1/2 K1/3 K1/4
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K K1/2 K1/3 K1/4
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what willhappen if k changes?
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a?
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a? We say these limit problems are of the form ∞0.
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a? We say these limit problems are of the form ∞0.
Similarly, if b(x) 0 and e(x) 0 as x a, we say these problems "lim b(x)e(x)" are of the form 00.
xa
Theorem: Let k > 0, then lim k1/x = 1.
In picture:x∞
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
0 1 KK1/2K1/3K1/4K K1/2 K1/3 K1/4
But instead of having a constant base k, what willhappen if k changes? In general, if b(x) ∞ and e(x) 0 as x a, what happens to b(x)e(x) as x a? We say these limit problems are of the form ∞0.
Similarly, if b(x) 0 and e(x) 0 as x a, we say these problems "lim b(x)e(x)" are of the form 00.
xa
Lastly, we have the 1∞ forms, i.e. the problems of "lim b(x)e(x)" where b(x) 1 and e(x) ∞.
xa
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) . (∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. x∞
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
We find the limit of Ln(x1/x) = first.Ln(x)x
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
this is ∞/∞ form so use the L'Hopital's Rule.
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
=
this is ∞/∞ form so use the L'Hopital's Rule.
lim [Ln(x)]'[x]'x∞
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
=
this is ∞/∞ form so use the L'Hopital's Rule.
lim [Ln(x)]'[x]'x∞
=
lim 1xx∞
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
=
this is ∞/∞ form so use the L'Hopital's Rule.
lim [Ln(x)]'[x]'x∞
=
lim 1xx∞
= 0.
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
=
this is ∞/∞ form so use the L'Hopital's Rule.
lim [Ln(x)]'[x]'x∞
=
lim 1xx∞
= 0.
So lim x1/x =x∞
lim eLn(x )1/x
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
=
this is ∞/∞ form so use the L'Hopital's Rule.
lim [Ln(x)]'[x]'x∞
=
lim 1xx∞
= 0.
So lim x1/x =x∞
lim eLn(x )1/x
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
=
this is ∞/∞ form so use the L'Hopital's Rule.
lim [Ln(x)]'[x]'x∞
=
lim 1xx∞
= 0.
So lim x1/x =x∞
lim eLn(x )1/x
(∞0, 00, 1∞)
0
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
These are the three indeterminate forms related tolimits of power functions b(x)e(x) .
We find their limits by first finding the limits of the logs of them i.e. Ln(b(x)e(x)) = e(x)*Ln(b(x)).
Example: A. Find lim x1/x. ( ∞0 form.)x∞
lim Ln(x1/x) =
We find the limit of Ln(x1/x) = first.Ln(x)x
lim Ln(x)xx∞ x∞
=
this is ∞/∞ form so use the L'Hopital's Rule.
lim [Ln(x)]'[x]'x∞
=
lim 1xx∞
= 0.
So lim x1/x =x∞
lim eLn(x )1/x = e0 = 1 .
0
(∞0, 00, 1∞)
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+ x0+
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+ x0+
lim Ln(x) x0+ x-1 =
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+ x0+
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+ x0+
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2
–x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+ x0+
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
–x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x
x0+
x0+
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 0
–x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x
x0+
x0+
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
–x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
–x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
–x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] lim x∞
lim x∞
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
–x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] lim x∞
lim x∞
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1=
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= =
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= lim
x∞[Ln(x + c) – Ln(x)]'
[x-1]'=
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= lim
x∞[Ln(x + c) – Ln(x)]'
[x-1]'=
lim x∞
1/(x + c) – 1/x-1/x2
=
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= lim
x∞[Ln(x + c) – Ln(x)]'
[x-1]'=
lim x∞
1/(x + c) – 1/x-1/x2
= lim x∞
x2 – x(x + c)-(x + 1)
=
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= lim
x∞[Ln(x + c) – Ln(x)]'
[x-1]'=
lim x∞
1/(x + c) – 1/x-1/x2
= lim x∞
x2 – x(x + c)-(x + 1)
= lim x∞
cxx + 1
=
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= lim
x∞[Ln(x + c) – Ln(x)]'
[x-1]'=
lim x∞
1/(x + c) – 1/x-1/x2
= lim x∞
x2 – x(x + c)-(x + 1)
= lim x∞
cxx + 1
= = c
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= lim
x∞[Ln(x + c) – Ln(x)]'
[x-1]'=
lim x∞
1/(x + c) – 1/x-1/x2
= lim x∞
x2 – x(x + c)-(x + 1)
= lim x∞
cxx + 1
= = c
Therefore lim (1 + c/x)x = ec.x∞
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
B. Find lim xx. (00 form)x0+
lim Ln(xx) = lim xLn(x) =x0+
So lim xx =x0+
lim eLn(x )x = 1
x0+
x0+
C. Find lim (1 + c/x)x where c is a constant. (1∞ form)x∞
Ln(1 + c/x)x = x*Ln[(x + c)/x] (∞*0 form)lim x∞
lim x∞
lim x∞
Ln(x + c) – Ln(x)
x-1= lim
x∞[Ln(x + c) – Ln(x)]'
[x-1]'=
lim x∞
1/(x + c) – 1/x-1/x2
= lim x∞
x2 – x(x + c)-(x + 1)
= lim x∞
cxx + 1
= = c
Therefore lim (1 + c/x)x = ec.x∞
lim Ln(x) x0+ x-1 =
L'Hopital
limx0+
x-1
-x-2 = 00
L'Hopital
–x
1
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞.
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x ( form, use L'Hopital's rule)
00
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x ( form, use L'Hopital's rule)
00
x0 [xsin(x)]'
[sin(x) – x]' = lim
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x ( form, use L'Hopital's rule)
00
x0 [xsin(x)]'
[sin(x) – x]' = lim
x0 sin(x)+ xcos(x)
cos(x) – 1= lim
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x ( form, use L'Hopital's rule)
00
x0 [xsin(x)]'
[sin(x) – x]' = lim
x0 sin(x)+ xcos(x)
cos(x) – 1= lim ( form, use L'Hopital's again)
00
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x ( form, use L'Hopital's rule)
00
x0 [xsin(x)]'
[sin(x) – x]' = lim
x0 sin(x)+ xcos(x)
cos(x) – 1= lim ( form, use L'Hopital's again)
00
x0 [sin(x)+ xcos(x)]'
[cos(x) – 1]'= lim
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x ( form, use L'Hopital's rule)
00
x0 [xsin(x)]'
[sin(x) – x]' = lim
x0 sin(x)+ xcos(x)
cos(x) – 1= lim ( form, use L'Hopital's again)
00
x0 [sin(x)+ xcos(x)]'
[cos(x) – 1]'= lim
x0 cos(x) + cos(x) – xsin(x)
-sin(x)= lim
D. Find lim ( )x0 x
L'Hopital's Rule II (∞0, 00, 1∞, ∞ – ∞ forms)
The last of the indeterminate forms are of the form +∞ – ∞ or -∞ + ∞. In these cases, we put the formulas into fractions first then take the limits.
1 sin(x)
1
lim( ) = x0 x
1 sin(x)
1 lim x0 xsin(x)
sin(x) – x ( form, use L'Hopital's rule)
00
x0 [xsin(x)]'
[sin(x) – x]' = lim
x0 sin(x)+ xcos(x)
cos(x) – 1= lim ( form, use L'Hopital's again)
00
x0 [sin(x)+ xcos(x)]'
[cos(x) – 1]'= lim
x0 cos(x) + cos(x) – xsin(x)
-sin(x)= lim = 0