Lim x → 4 1 + 1 4 x = 2 given ε > 0, we need δ such that if 0 < |x − 4| < δ, then 1 + 1 4 x − 2 . but 1 + 1 4 x − 2 < ε ⇔ 1 4 x − 1 < ε ⇔ 1 4 |x − 4| < ε ⇔ |x − 4| < . so if we choose δ = then 0 < |x − 4| < δ ⇒ 1 + 1 4 x − 2 < ε. thus, lim x → 4 1 + 1 4 x = 2 by the definition of a limit.

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation:

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !

this "question" looks more like a project is there any other info you can give me about this. it could be too complex for someone else to do on this website