English For Socializing [PDF]

Zitiervorschau

English for Socializing Task 5 p. 25 Sport Skiing Doing yoga Keeping fit Playing football

Relaxing at home Listening to music Reading Watching television Cooking

Going out Eating out Going to the cinema Shopping Going to concerts

Jogging Doing Sky diving

Having a bath Playing with family

Going to the exhibition Visiting new town

Task 6 p. 25 1. 2. 3. 4. 5. 6. 7. 8.

Go Playing Have/Done Do/Go I’ve been playing Did Do Do

Task 10 p. 28 1–e–C 2–g–F 3–a–G 4–f–A 5–b–D 6–c–B 7–d–E

Text p. 30 Tip #1: Think in English. First time that will seem to be some king of wired thing, but pleasure and skills comes with practice Tip #2: Write notes in English.

Tip #3: Listen podcasts or audiobooks in addition to video (of course in English), later you will find yourself listening it in x2 speed with the excellent understanding.

SpeakOut Task 5 A p. 17 You said just READ question.

Task 5 B p. 17 1. 2. 3. 4. 5. 6.

Something Happy Love Thing Drive Good

Task 5 C p. 17 Name – Alex, Crash fact about me: Just love unusual not ordinary crazy things. Prefer progress instead of regress, that’s why I’m always ready to absorb new information with the extremely high level of energy and passion. My lifestyle? Don’t have it, know why? I’m the owner of my life and every of my choices depends on me, my lifestyle is choice too. I’m flexible in my choices and that’s cool for me. But of course I have my own values, which are important for me and they play not the bottom role in my choices. What about my greatest achievement? Few years ago I took my life in my hands. Now I’m responsible for my fantastic life. Proud of it.

Завдання 1. Дослідити задані функції f (x) на неперервність і встановити характер точок розриву. Зробити схематичний рисунок. 2 x2 +1 , x ≤ 0 ln |x−ⅇ|, 0< x ≤2 ⅇ а) f ( x )= πx sin , x >2 ⅇ 4ⅇ

{

( )

D ( y )=(−∞ ; 0 ) ∪ ( 0 ; ⅇ ) ∪ ( ⅇ ; 2 ⅇ ) v ( 2e ;+∞ ) f ( 0 )=1=f ( 2 ⅇ )=1 x 1=0 , x 2=ⅇ , x 3=2 ⅇ−¿ можливі точки розриву.

а)

x 1=0 →

lim x →0−¿ f ( x ) =

¿ lim

¿¿

2

x →0−¿ (2 x +1 )=1¿

lim x→ 0+¿ f ( x ) =

x →O+¿ln |x−ⅇ|→

¿ lim

lim

x →0+¿ |ⅇ −x|=lnⅇ =1=f ( 0 ) ⇒x −¿ ¿ 0

¿¿ ¿¿

точка неп. f(x), розриву немає.

lim f ( x )= lim ln ( ⅇ−x )=|ln ( ⅇ−x ) → {−∞ } , x−ⅇ →−0|={−∞ } б) x 2=ⅇ , x→ ⅇ−0 x→ e−0 lim f ( x )= lim ln ( ⅇ−x )=|ln ( ⅇ−x ) → {−∞ } , x−ⅇ →+ 0|={−∞ }

x→ ⅇ+0

x →e+0

Точка x 2=ⅇ−¿ точка розриву 2 – го роду.

в) x 3=ⅇ , lim f ( x ) = lim sin x →2 ⅇ+0

x →2 ⅇ+0

( 4πxx )=sin 24πⅇⅇ =sin π2 =1

lim f ( x ) = lim f ( x )=1=f ( 2 ⅇ )=¿ x 3=2 ⅇ−точка неперервності f(x).

x→ 2 ⅇ−0

x →2 ⅇ+0

Відповідь: 1. Tочка x=ⅇ−¿ точка розриву 2-го роду f(x) 2. ∀ x ∈ D ( y ) : x ≠ ⅇ , f ( x )−¿ неперервна.

( sin x )2 n б) f ( x )=nlim →∞ f ( 0 )=lim ( sin 0 )2 n=0 , f ( πk ) =lim ( sin πk )2 π =0 n→∞

f

(

n→ ∞

π π ±2 πk =lim sin ± +2 πk 2 2 n →∞

( (

)

{

Нехай x ∈ R ∖ πk ; ±

2n

)) =lim ⁡¿

π +2 πk =¿ sin x=q ,|q|