Name: Bekenov Kuandyk
Assignment 1

1. Prove that fx=2x+5 is one-to-one function. (10 points)

Let make an assumption tha fx1=fx2t and then prove that x1=x2.

fx1=fx2=2x1+5=2x2+5→x1=x2 ,

Then fx=2x+5 is one-to-one function

2. B
B
A
A
Let f: A→B, as given below. Is f a one-to-one function? Please explain why or why not. (10 points)

f f 5
5
5
5
1
1

3
3
1
1

6
6
2
2

2
2
4
4
6
6

7
7
4
4

8
8
3
3

F is not one-to-one function, because f(1)=1 and f(2)=1.

3. The modulo function (a mod n or a modulo n) maps every positive integer number to the remainder of the division of a/n. For example, the expression 22 mod 5 would evaluate to 2 since 22 divided by 5 is 4 with a remainder of 2. The expression 10 mod 5 would resolve to 0 since 10 is divisible by 5 and there is not a remainder. a. If n is fixed as 5, is this function one-to-one? (5 points)
F(n mod 5) is not one-to-one function, because f(n mod 5)= between 0 and 4. It means that we have the same image of F for different n (integer). b. List five numbers that have the exact same image. (5 points)
12, 17, 22, 27, 32

4. Find limn→∞n2-1n3+n (6 points)

limn→∞n2-1n3+n=limn→∞n2n3-1n3n3n3+nn3=0-01+0=0

5. Find limx→∞x1001-xx1000+x (6 points)

limx→∞x1001-xx1000+x=limx→∞1-1x10001x+1x1000=1-00+0=∞

6. Find limx→∞8x6+4x4-3x2x6+x5-7 (6 points) limx→∞8x6+4x4-3x2x6+x5-7=limx→∞8+4x2-3x52+1x-7x6=8+0-02+0-0=4 7. Find lim2x99+x+1,000,000x100+2,000,000 as x→0 (6 points)

limx→02x99+x+1000000x100+2000000=0+0+10000000+2000000=0.5

8. Prove that fx=x2+7 is a continuous function. (8 points)

Let prove that fx+∆x-fx→0 when ∆x→0 then: x+∆x2+7-x2+7=x2+2x∆x+∆x2+7-x2-7-=∆x2x+∆x=0, it means fx= x2+7 is a continuous function.

9. Find derivatives of the following