Question

If lim x-> infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b

Answer 1

lim(x->∞)[(x^2+1)/x+1)-ax-b]=0
i.e. lim(x->∞)[(x^2+2x+1-2x)/x+1)-ax-b]=0
i.e. lim(x->∞)[((x+1)^2-2x)/x+1)-ax-b]=0
i.e. lim(x->∞)[((x+1)^2-2x)/x+1)]=lim(x->∞)[a...
i.e. lim(x->∞)[((x+1)^2)/x+1)- 2x/x+1]=lim(x->∞)[ax+b]
i.e. lim(x->∞)[(x+1)- 2x/x+1]=lim(x->∞)[ax+b]
i.e. lim(x->∞)[(x+1)- (2x+2-2)/x+1]=lim(x->∞)[ax+b]
i.e. lim(x->∞)[(x+1)-2 +2/x+1]=lim(x->∞)[ax+b]
i.e. lim(x->∞)[(x-1 +2/(x+1)]=lim(x->∞)[ax+b]
taking limit on both sides.
i.e. lim(x->∞)x -1 +0= alim(x->∞)x+b
thus a=1 and b=-1

Answer 2

Answer:

a=1

b=-1

Step-by-step explanation:

lim(x->∞)[(x^2+1)/x+1)-ax-b]=0

i.e. lim(x->∞)[(x^2+2x+1-2x)/x+1)-ax-b]=0

i.e. lim(x->∞)[((x+1)^2-2x)/x+1)-ax-b]=0

i.e. lim(x->∞)[((x+1)^2-2x)/x+1)]=lim(x->∞)[a...

i.e. lim(x->∞)[((x+1)^2)/x+1)- 2x/x+1]=lim(x->∞)[ax+b]

i.e. lim(x->∞)[(x+1)- 2x/x+1]=lim(x->∞)[ax+b]

i.e. lim(x->∞)[(x+1)- (2x+2-2)/x+1]=lim(x->∞)[ax+b]

i.e. lim(x->∞)[(x+1)-2 +2/x+1]=lim(x->∞)[ax+b]

i.e. lim(x->∞)[(x-1 +2/(x+1)]=lim(x->∞)[ax+b]

taking limit on both sides.

i.e. lim(x->∞)x -1 +0= alim(x->∞)x+b

thus a=1 and b=-1

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