Evaluate in Software
Evaluate Painting Quick Response Code In None Using Barcode creation for Software Control to generate, create QR image in Software applications. Recognizing Quick Response Code In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. x +
QR Encoder In Visual C#.NET Using Barcode creation for .NET framework Control to generate, create QR Code 2d barcode image in VS .NET applications. Printing QR Code ISO/IEC18004 In .NET Using Barcode generation for ASP.NET Control to generate, create QRCode image in ASP.NET applications. x4 ex
QR Code Creator In Visual Studio .NET Using Barcode generation for VS .NET Control to generate, create QR Code 2d barcode image in Visual Studio .NET applications. Making QR In VB.NET Using Barcode creator for VS .NET Control to generate, create QR image in VS .NET applications. SOLUTION We rst notice that both the numerator and the denominator tend to + as x + Thus the quotient is indeterminate at + of the form + / + Therefore the new version of l H pital s Rule applies and our limit equals 4x 3 x + ex Again the numerator and denominator tend to + as x + , so we once more apply l H pital s Rule The limit equals lim 12x 2 = 0 x + ex We must apply l H pital s Rule two more times We rst obtain lim 24x x + ex lim Draw Code 3/9 In None Using Barcode drawer for Software Control to generate, create USS Code 39 image in Software applications. ANSI/AIM Code 128 Generator In None Using Barcode generator for Software Control to generate, create Code 128 Code Set A image in Software applications. CHAPTER 5 Indeterminate Forms
Bar Code Generator In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. Drawing UCC.EAN  128 In None Using Barcode printer for Software Control to generate, create GS1128 image in Software applications. and then 24 x + ex lim We conclude that x4 = 0 x + ex ex You Try It: Evaluate the limit limx + x ln x You Try It: Evaluate the limit limx x 4 ex lim EXAMPLE 55 Encode EAN13 In None Using Barcode creation for Software Control to generate, create EAN13 image in Software applications. Creating Barcode In None Using Barcode printer for Software Control to generate, create barcode image in Software applications. Evaluate the limit
Encode GS1  12 In None Using Barcode creation for Software Control to generate, create UPCE image in Software applications. Barcode Decoder In VB.NET Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in VS .NET applications. x Encoding ANSI/AIM Code 39 In VB.NET Using Barcode maker for .NET framework Control to generate, create Code 3/9 image in .NET framework applications. Bar Code Decoder In C# Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in Visual Studio .NET applications. sin(2/x) sin(5/x) Paint UCC.EAN  128 In C#.NET Using Barcode printer for Visual Studio .NET Control to generate, create UCC128 image in .NET framework applications. Barcode Encoder In VB.NET Using Barcode encoder for Visual Studio .NET Control to generate, create bar code image in Visual Studio .NET applications. SOLUTION We note that both numerator and denominator tend to 0, so the quotient is indeterminate at of the form 0/0 We may therefore apply l H pital s Rule Our limit equals ( 2/x 2 ) cos(2/x) x ( 5/x 2 ) cos(5/x) lim This in turn simpli es to 2 2 cos(2/x) = x 5 cos(5/x) 5 lim l H pital s Rule also applies to onesided limits Here is an example EXAMPLE 56 Decoding Code 39 Full ASCII In .NET Using Barcode scanner for .NET Control to read, scan read, scan image in Visual Studio .NET applications. Print Code 128C In None Using Barcode creator for Font Control to generate, create Code 128A image in Font applications. Evaluate the limit
x 0+
x SOLUTION Both numerator and denominator tend to zero so the quotient is indeterminate at 0 of the form 0/0 We may apply l H pital s Rule; differentiating numerator and denominator, we nd that the limit equals [cos x] (1/2)x 1/2 lim = lim cos x 1/2 + + (1/2)x x 0 x 0 = 1 You Try It: How can we apply l H pital s Rule to evaluate limx 0+ x ln x Indeterminate Forms
Other Indeterminate Forms
521 INTRODUCTION
By using some algebraic manipulations, we can reduce a variety of indeterminate limits to expressions which can be treated by l H pital s Rule We explore some of these techniques in this section WRITING A PRODUCT AS A QUOTIENT
The technique of the rst example is a simple one, but it is used frequently EXAMPLE 57
Evaluate the limit
x lim x 2 e3x
SOLUTION Notice that x 2 + while e3x 0 So the limit is indeterminate of the form 0 We rewrite the limit as x e 3x
Now both numerator and denominator tend to in nity and we may apply l H pital s Rule The result is that the limit equals 2x x 3e 3x lim Again the numerator and denominator both tend to in nity so we apply l H pital s Rule to obtain: x It is clear that the limit of this last expression is zero We conclude that
x You Try It: Evaluate the limit limx + e
THE USE OF THE LOGARITHM
The natural logarithm can be used to reduce an expression involving exponentials to one involving a product or a quotient AM FL Y
lim x2 lim 2 9e 3x lim x e3x = 0
x CHAPTER 5 Indeterminate Forms
EXAMPLE 58
Evaluate the limit
x 0+
lim x x
SOLUTION We study the limit of f (x) = x x by considering ln f (x) = x ln x We rewrite this as ln x lim ln f (x) = lim x 0+ 1/x x 0+ Both numerator and denominator tend to , so the quotient is indeterminate of the form / Thus l H pital s Rule applies The limit equals x 0+
1/x = lim x = 0 1/x 2 x 0+
Now the only way that ln f (x) can tend to zero is if f (x) = x x tends to 1 We conclude that
x 0+
lim x x = 1
EXAMPLE 59
Evaluate the limit
x 0 lim (1 + x 2 )ln x 
SOLUTION Let f (x) = (1 + x 2 )ln x and consider ln f (x) = ln x ln(1 + x 2 ) This expression is indeterminate of the form 0 We rewrite it as ln(1 + x 2 ) , x 0 1/ ln x lim so that both the numerator and denominator tend to 0 So l H pital s Rule applies and we have lim ln f (x) = lim 2x/(1 + x 2 ) 1/[x ln2 (x)] = lim 2x 2 ln2 (x) (1 + x 2 ) The numerator tends to 0 (see Example 53) and the denominator tends to 1 Thus
lim ln f (x) = 0 Indeterminate Forms
But the only way that ln f (x) can tend to zero is if f (x) tends to 1 We conclude that
lim (1 + x 2 )ln x = 1
You Try It: Evaluate the limit limx 0+ (1/x)x You Try It: Evaluate the limit limx 0+ (1 + x)1/x In fact this limit gives an important way to de ne Euler s constant e (see Sections 19 and 623)

