Question

1.

a) Prove: if $\lim_{x\rightarrow a}f(x) = L$ and $\lim_{x\rightarrow a}g(x)=M$ , then $\lim_{x\rightarrow a}(f(x)\cdot g(x))= L\cdot M$

b) State the converse above, and find a counterexample to the converse above.

Answer 1

X+ a 2a x→a (1) - (2) Mta nya lim lim (919-M)= neta (a) lim final, lim g(x)=M Claim: lim (61a). g(x)) = 1.M Proof: lim f(x)= L = lim (time) - 2) = 0 g(x)=M uring timest definition. fet Ero be given tren 36,>0,8270 042-01 <0,al blm) -LILLE o<lx-akda ²1 fix)-MILIE Choone 8= min {0,,d2} so whenever os lu-al< o we get | Cbla (W)- L) }{(lm) – Mj}] | 6(x)-21 lg 10-ml 1 <de de no Nta До 166(X)=2169169-M)-01 Et line (6180-L) 1.913-M) = 0 (3) Now consider (61m)-L) (910)-M)= () g(x) -L 91x) -M 61x+LM bland g(x)=(6(X)- L) (91%) –M) + 9(x) Mf( so lim (fexiglas) = lim (bini-ll 18 (*)-M) + L lin 9 (1) + M lim f(x)-LP uning (3) lim (broni grm). 0 + L M + ML-LM= L M lim ( fial 9 623 ) = lim fing. lim 9 (0) - LM Mta Nya aya 21a M nta nta naa CScanned with CamScanner Q.E.D.

(6) Converse of a lim (bm).g103) = L.M 2a lim fial=L and dim g(x)=M Na Counter example: a=0 let blan) = x, 919) blagog (n)=sim 1 x on IR-103 슬 lim no wito = 1 . x nto lim ne = 0 no but lim fen) no and lim 1 / 4 does not exist no so lim 2-4a (b1w). 3IN) ) = L-M lim f(x)=L f (x)=L and lim 9 (2)=M 2-4a CScanned with damScannem ya