Mixing
Find $alpha$,$beta$ if $lim_{x→∞}[ sqrt {ax^2+2bx+c} - alpha x -beta] = 0$
Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = frac {-b±sqrt {b^2-ac}}{a} = frac {-b±sqrt D}{a}$$
Hence;
$$sqrt {ax^2+2bx+c} = sqrt {(x+frac {b-√D}{a})(x+frac {b+√D}{a})}$$
$$=xsqrt {(1+frac {b-√D}{ax})(1+frac {b+√D}{ax})}$$
For large values of $x$ we may apply the binomial approximation, so that;
$$sqrt {ax^2+2bx+c} ≈ x(1+frac {b-√D}{2ax})(1+frac {b+√D}{2ax})$$
$$=x + frac {b}{a} + frac {c}{4ax}$$
As $x→∞$ the final term in the above expression vanishes. Hence;
$$lim_{x→∞}[ sqrt {ax^2+2bx+c} - alpha x - beta ] = 0,$$
gives;
$$x+frac {b}{a} - alpha x - beta = 0,$$
or;
$$(1-alpha)x + (frac {b}{a} - beta) = 0$$
As $x→∞$, $1-alpha$ must be $0$ for the former term to vanish, hence,
$$alpha = 1, beta = frac {b}{a}$$
But I doubt it is hardly correct. Is there any better method for the problem?
limits
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
add a comment |
Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = frac {-b±sqrt {b^2-ac}}{a} = frac {-b±sqrt D}{a}$$
Hence;
$$sqrt {ax^2+2bx+c} = sqrt {(x+frac {b-√D}{a})(x+frac {b+√D}{a})}$$
$$=xsqrt {(1+frac {b-√D}{ax})(1+frac {b+√D}{ax})}$$
For large values of $x$ we may apply the binomial approximation, so that;
$$sqrt {ax^2+2bx+c} ≈ x(1+frac {b-√D}{2ax})(1+frac {b+√D}{2ax})$$
$$=x + frac {b}{a} + frac {c}{4ax}$$
As $x→∞$ the final term in the above expression vanishes. Hence;
$$lim_{x→∞}[ sqrt {ax^2+2bx+c} - alpha x - beta ] = 0,$$
gives;
$$x+frac {b}{a} - alpha x - beta = 0,$$
or;
$$(1-alpha)x + (frac {b}{a} - beta) = 0$$
As $x→∞$, $1-alpha$ must be $0$ for the former term to vanish, hence,
$$alpha = 1, beta = frac {b}{a}$$
But I doubt it is hardly correct. Is there any better method for the problem?
limits
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
add a comment |
Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = frac {-b±sqrt {b^2-ac}}{a} = frac {-b±sqrt D}{a}$$
Hence;
$$sqrt {ax^2+2bx+c} = sqrt {(x+frac {b-√D}{a})(x+frac {b+√D}{a})}$$
$$=xsqrt {(1+frac {b-√D}{ax})(1+frac {b+√D}{ax})}$$
For large values of $x$ we may apply the binomial approximation, so that;
$$sqrt {ax^2+2bx+c} ≈ x(1+frac {b-√D}{2ax})(1+frac {b+√D}{2ax})$$
$$=x + frac {b}{a} + frac {c}{4ax}$$
As $x→∞$ the final term in the above expression vanishes. Hence;
$$lim_{x→∞}[ sqrt {ax^2+2bx+c} - alpha x - beta ] = 0,$$
gives;
$$x+frac {b}{a} - alpha x - beta = 0,$$
or;
$$(1-alpha)x + (frac {b}{a} - beta) = 0$$
As $x→∞$, $1-alpha$ must be $0$ for the former term to vanish, hence,
$$alpha = 1, beta = frac {b}{a}$$
But I doubt it is hardly correct. Is there any better method for the problem?
limits
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
Here is my approach.
Consider;
$$ax^2 + 2bx + c = 0$$
or;
$$ x_{±} = frac {-b±sqrt {b^2-ac}}{a} = frac {-b±sqrt D}{a}$$
Hence;
$$sqrt {ax^2+2bx+c} = sqrt {(x+frac {b-√D}{a})(x+frac {b+√D}{a})}$$
$$=xsqrt {(1+frac {b-√D}{ax})(1+frac {b+√D}{ax})}$$
For large values of $x$ we may apply the binomial approximation, so that;
$$sqrt {ax^2+2bx+c} ≈ x(1+frac {b-√D}{2ax})(1+frac {b+√D}{2ax})$$
$$=x + frac {b}{a} + frac {c}{4ax}$$
As $x→∞$ the final term in the above expression vanishes. Hence;
$$lim_{x→∞}[ sqrt {ax^2+2bx+c} - alpha x - beta ] = 0,$$
gives;
$$x+frac {b}{a} - alpha x - beta = 0,$$
or;
$$(1-alpha)x + (frac {b}{a} - beta) = 0$$
As $x→∞$, $1-alpha$ must be $0$ for the former term to vanish, hence,
$$alpha = 1, beta = frac {b}{a}$$
But I doubt it is hardly correct. Is there any better method for the problem?
limits
limits
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
asked Nov 21 '18 at 5:35
Awe Kumar Jha
38613
38613
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Set $1/x=h$ to find $$lim_{hto0^+}dfrac{sqrt{a+bh+ch^2}-(alpha+beta h)}h$$
$$=lim_{hto0^+}dfrac{(a+bh+ch^2)-(alpha+beta h)^2}hcdotlim_{hto0^+}dfrac1{sqrt{a+bh+ch^2}-(alpha+beta h)}$$
$(a+bh+ch^2)-(alpha+beta h)^2=a-alpha^2+h(b-2alphabeta)+h^2(c-beta^2)$
As the denominator $to0,$ $$a-alpha^2=0$$
as the denominator is $O(h)$ $$b-2alphabeta=0$$
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
add a comment |
Another approach could be Taylor series
$$sqrt{a x^2+2 b x+c}=x sqrt{a+frac{2 b}{x}+frac{c}{x^2} }$$ So, for large $x$,
$$sqrt{a x^2+2 b x+c}=sqrt{a} x+frac{b}{sqrt{a}}+frac{a c-b^2}{2 a^{3/2}
x}+Oleft(frac{1}{x^2}right)$$
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
I like THIS approach! (+1)
– Robert Z
Nov 21 '18 at 6:39
add a comment |
You want $$ lim _{xto infty }ax^2+2bx+c -(alpha x +beta )^2=0$$
That implies $$ a= alpha ^2, c=beta ^2, b=alpha beta$$
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
$b=2alphabeta$ right? and $c-beta^2$ needs to be finite
– lab bhattacharjee
Nov 21 '18 at 6:35
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Set $1/x=h$ to find $$lim_{hto0^+}dfrac{sqrt{a+bh+ch^2}-(alpha+beta h)}h$$
$$=lim_{hto0^+}dfrac{(a+bh+ch^2)-(alpha+beta h)^2}hcdotlim_{hto0^+}dfrac1{sqrt{a+bh+ch^2}-(alpha+beta h)}$$
$(a+bh+ch^2)-(alpha+beta h)^2=a-alpha^2+h(b-2alphabeta)+h^2(c-beta^2)$
As the denominator $to0,$ $$a-alpha^2=0$$
as the denominator is $O(h)$ $$b-2alphabeta=0$$
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
add a comment |
Set $1/x=h$ to find $$lim_{hto0^+}dfrac{sqrt{a+bh+ch^2}-(alpha+beta h)}h$$
$$=lim_{hto0^+}dfrac{(a+bh+ch^2)-(alpha+beta h)^2}hcdotlim_{hto0^+}dfrac1{sqrt{a+bh+ch^2}-(alpha+beta h)}$$
$(a+bh+ch^2)-(alpha+beta h)^2=a-alpha^2+h(b-2alphabeta)+h^2(c-beta^2)$
As the denominator $to0,$ $$a-alpha^2=0$$
as the denominator is $O(h)$ $$b-2alphabeta=0$$
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
add a comment |
Set $1/x=h$ to find $$lim_{hto0^+}dfrac{sqrt{a+bh+ch^2}-(alpha+beta h)}h$$
$$=lim_{hto0^+}dfrac{(a+bh+ch^2)-(alpha+beta h)^2}hcdotlim_{hto0^+}dfrac1{sqrt{a+bh+ch^2}-(alpha+beta h)}$$
$(a+bh+ch^2)-(alpha+beta h)^2=a-alpha^2+h(b-2alphabeta)+h^2(c-beta^2)$
As the denominator $to0,$ $$a-alpha^2=0$$
as the denominator is $O(h)$ $$b-2alphabeta=0$$
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
Set $1/x=h$ to find $$lim_{hto0^+}dfrac{sqrt{a+bh+ch^2}-(alpha+beta h)}h$$
$$=lim_{hto0^+}dfrac{(a+bh+ch^2)-(alpha+beta h)^2}hcdotlim_{hto0^+}dfrac1{sqrt{a+bh+ch^2}-(alpha+beta h)}$$
$(a+bh+ch^2)-(alpha+beta h)^2=a-alpha^2+h(b-2alphabeta)+h^2(c-beta^2)$
As the denominator $to0,$ $$a-alpha^2=0$$
as the denominator is $O(h)$ $$b-2alphabeta=0$$
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
answered Nov 21 '18 at 5:42
lab bhattacharjee
223k15156274
223k15156274
add a comment |
add a comment |
Another approach could be Taylor series
$$sqrt{a x^2+2 b x+c}=x sqrt{a+frac{2 b}{x}+frac{c}{x^2} }$$ So, for large $x$,
$$sqrt{a x^2+2 b x+c}=sqrt{a} x+frac{b}{sqrt{a}}+frac{a c-b^2}{2 a^{3/2}
x}+Oleft(frac{1}{x^2}right)$$
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
I like THIS approach! (+1)
– Robert Z
Nov 21 '18 at 6:39
add a comment |
Another approach could be Taylor series
$$sqrt{a x^2+2 b x+c}=x sqrt{a+frac{2 b}{x}+frac{c}{x^2} }$$ So, for large $x$,
$$sqrt{a x^2+2 b x+c}=sqrt{a} x+frac{b}{sqrt{a}}+frac{a c-b^2}{2 a^{3/2}
x}+Oleft(frac{1}{x^2}right)$$
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
I like THIS approach! (+1)
– Robert Z
Nov 21 '18 at 6:39
add a comment |
Another approach could be Taylor series
$$sqrt{a x^2+2 b x+c}=x sqrt{a+frac{2 b}{x}+frac{c}{x^2} }$$ So, for large $x$,
$$sqrt{a x^2+2 b x+c}=sqrt{a} x+frac{b}{sqrt{a}}+frac{a c-b^2}{2 a^{3/2}
x}+Oleft(frac{1}{x^2}right)$$
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
Another approach could be Taylor series
$$sqrt{a x^2+2 b x+c}=x sqrt{a+frac{2 b}{x}+frac{c}{x^2} }$$ So, for large $x$,
$$sqrt{a x^2+2 b x+c}=sqrt{a} x+frac{b}{sqrt{a}}+frac{a c-b^2}{2 a^{3/2}
x}+Oleft(frac{1}{x^2}right)$$
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
answered Nov 21 '18 at 6:18
Claude Leibovici
119k1157132
119k1157132
I like THIS approach! (+1)
– Robert Z
Nov 21 '18 at 6:39
add a comment |
I like THIS approach! (+1)
– Robert Z
Nov 21 '18 at 6:39
I like THIS approach! (+1)
– Robert Z
Nov 21 '18 at 6:39
I like THIS approach! (+1)
– Robert Z
Nov 21 '18 at 6:39
add a comment |
You want $$ lim _{xto infty }ax^2+2bx+c -(alpha x +beta )^2=0$$
That implies $$ a= alpha ^2, c=beta ^2, b=alpha beta$$
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
$b=2alphabeta$ right? and $c-beta^2$ needs to be finite
– lab bhattacharjee
Nov 21 '18 at 6:35
add a comment |
You want $$ lim _{xto infty }ax^2+2bx+c -(alpha x +beta )^2=0$$
That implies $$ a= alpha ^2, c=beta ^2, b=alpha beta$$
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
$b=2alphabeta$ right? and $c-beta^2$ needs to be finite
– lab bhattacharjee
Nov 21 '18 at 6:35
add a comment |
You want $$ lim _{xto infty }ax^2+2bx+c -(alpha x +beta )^2=0$$
That implies $$ a= alpha ^2, c=beta ^2, b=alpha beta$$
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
You want $$ lim _{xto infty }ax^2+2bx+c -(alpha x +beta )^2=0$$
That implies $$ a= alpha ^2, c=beta ^2, b=alpha beta$$
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
answered Nov 21 '18 at 6:18
Mohammad Riazi-Kermani
41k42059
41k42059
$b=2alphabeta$ right? and $c-beta^2$ needs to be finite
– lab bhattacharjee
Nov 21 '18 at 6:35
add a comment |
$b=2alphabeta$ right? and $c-beta^2$ needs to be finite
– lab bhattacharjee
Nov 21 '18 at 6:35
$b=2alphabeta$ right? and $c-beta^2$ needs to be finite
– lab bhattacharjee
Nov 21 '18 at 6:35
$b=2alphabeta$ right? and $c-beta^2$ needs to be finite
– lab bhattacharjee
Nov 21 '18 at 6:35
add a comment |
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