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Since F x; y x 4 y 2 sin 1 y= x 6 x4 y2 sin 1 y=x 6 F x; y , the result follows from Problem 6.25 with p 6. It can of course also be shown by direct di erentiation.

The functi ons in the next example exhibit a different behavior. As x 0, the function oscillates an infinite number of times. Mathematica returns the limit as an Interval object. Interval[{min, max}] represents the range of values between min and max.

Limit[Sin[1/x], x 0] Interval[{ 1, 1}] Limit[Tan[1/x], x 0] Interval[{ , }]

1 (1+ Log[2]) 3

6.27. Prove that Y f x at g x at satis es @2 Y=@t2 a2 @2 Y=@x2 , where f and g are assumed to be at least twice di erentiable and a is any constant.

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x 0

x 0

Limit[(1 + Sin[x])Cot[2x], x 0]

Let u x at; v x at so that Y f u g v . @Y @Y @u @Y @v a f 0 u ag 0 v ; @t @u @t @v @t Then if f 0 u df =du, g 0 v dg=dv, @Y @Y @u @Y @v f 0 u g 0 v @x @x @x @v @x

1/x Limit[(Exp[x]+ x) , x ]

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1/x Limit[(Exp[x]+ x) , x ] 1

8.5 Compute lim(2 x )

By further di erentiation, using the notation f 00 u d 2 f =du2 , g 00 v d 2 g=dv2 , we have 1 @2 Y @Yt @Yt @u @Yt @v @ @ fa f 0 u a g 0 v g a fa f 0 u a g 0 v g a @t @u @t @v @t @u @v @t2 a2 f 00 u a2 g 00 v @2 Y @Yx @Yx @u @Yx @v @ @ f f 0 u g 0 v g f f 0 u g 0 v g @x @u @x @v @x @u @v @x2 f 00 u g 00 v Then from (1) and (2), @2 Y=@t2 a2 @2 Y=@x2 .

x 1

tan( x ) 2

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CHAP. 6]

Limit[(2 x)Tan[ 2 x], x 1] 2/

8.6 If p dollars is compounded n times per year at an annual interest rate of r, the money will be worth nt r p 1 + dollars after t years. How much will the money be worth after t years if it is compounded n continuously (n )

8.7 The derivative of a function is defined to be lim h 0 derivative of f ( x ) = ln x + x 5 + sin x .

Method 1: Di erentiate the given equations with respect to x, considering u and v as functions of x and y. Then 1 @u 1 12v ; @x 1 8uv 2u @u @v 3 @x @x 2 @u @v 4v 1 @x @x

h 0

- 211 -

f[x_]= Log[x]+ x 5 + Sin[x]; Limit f[x + h] 2 f[x]+ f[x h], h 0 h2 12 + 20 x3 Sin[x] x

There are several ways derivatives can be computed in Mathematica. Each has its advantages and disadvantages, so the proper choice for a particular situation must be determined.

Solving,

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