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// Demonstrate dynamic initialization using System; class DynInit static void // Length double s1 double s2 { Main() { of sides = 40; = 50;

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

The acceleration boundary conditions give: 8p h 16p 2 4p q p = -C3 + sin 2 3 b 3 q = b b 9b 2

// Dynamically initialize hypot double hypot = MathSqrt( (s1 * s1) + (s2 * s2) ); ConsoleWrite("Hypotenuse of triangle with sides " + s1 + " by " + s2 + " is "); ConsoleWriteLine("{0:####}", hypot); } }

Here is the output:

3:

9h C3 = 2p The velocity boundary conditions give: 2h C2 9h 4p 4p q p = + cos 3b 3 b 3 q = b b b 2p

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Here, three local variables s1, s2, and hypot are declared The first two, s1 and s2, are initialized by constants However, hypot is initialized dynamically to the length of the hypotenuse Notice that the initialization involves calling MathSqrt( ) As explained, you can use any expression that is valid at the point of the initialization Since a call to MathSqrt( ) (or any other library method) is valid at this point, it can be used in the initialization of hypot The key point here is that the initialization expression can use any element valid at the time of the initialization, including calls to methods, other variables, or literals

x 0

As explained, in C# all variables must be declared Normally, a declaration includes the type of the variable, such as int or bool, followed by the name of the variable However, beginning with C# 30, it is possible to let the compiler determine the type of a local variable based on the value used to initialize it This is called an implicitly typed variable An implicitly typed variable is declared using the keyword var, and it must be initialized The compiler uses the type of the initializer to determine the type of the variable Here is an example:

C2 = 2 h The displacement boundary conditions give: b q 8 - 9h sin 4p q + p =0 C1 + 2 h b 3 b 3 2p b q =

var e = 27183;

Because e is initialized with a floating-point literal (whose type is double by default), the type of e is double Had e been declared like this:

Then q - b h - h + 9h + 2 h 8 - 9h sin 4p q + p y= 4 2p 2p b 2p 3 b 3 q 9 4 4p q p sin = h + 2 + 3 b 3 b 2p p When q = b , the rise is 2 4 y = h 1 + p Hence, 3 9 y2 = y - y1 = + h 4 2p Finally, from the relationship y = y1 + y2 = p h 2(p + 4) b 8 (317) h 2

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

var e = 27183F;

then e would have the type float, instead The following program demonstrates implicitly typed variables It reworks the program shown in the preceding section so that all variables are implicitly typed

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