Value at a limit

[Arbu]

The function y= sqrt(x^2 +3x) has a gradient of 1 as x tends to infinity. So the function approaches a line y=x+c. How do you calculate c?

[Guest]

For very large x, [tex]x^2[/tex] is much larger than [tex]3x[/tex] so that [tex]\sqrt{x^2+ 3x}[/tex] approximates [tex]\sqrt{x^2}= |x|[/tex] which, since x is positive, is x. The "c" is 0.

[Arbu]

Well let's work out sqrt(x^2+3x) - x to four decimal places for various values of x:

2: 1.1623
5: 1.3246
20: 1.4476
100: 1.4889
1000: 1.4989
100,000: 1.5000

Looks a lot like the answer is 1.5 to me, not 0.

But how do you show this mathematically? Something to do with mathematical expansions I think.

[Arbu]

Rewrite the function as y = x(1+3/x)^0.5
By the binomial theorem for non positive integer exponents (https://en.wikipedia.org/wiki/Binomial_theorem#Newton's_generalized_binomial_theorem) y = x(1+3/2x-9/8x^2...)
So as x tends to infinity, y tends to x + 1.5. So c = 1.5

[Guest]

What I would do:
Write [tex]\sqrt{x^2+ 3x}- (x+ k)[/tex]
"Rationalize the numerator" by multiplying numerator and denominator by [tex]\sqrt{x^2+ 3x}+ (x+ k)[/tex]:

[tex]\frac{(\sqrt{x^2+ 3x})^2- (x+ k)^2}{\sqrt{x^2+ 3x}+ (x+ k)}= \frac{x^2+ 3x- x^2- 2kx+ k^2}{\sqrt{x^2+ 3x}+ (x+ k)}[/tex]
[tex]\frac{3x- 2kx+ k^2}{\sqrt{x^2+ 3x}+ (x+ k)}[/tex]

Divide both numerator and denominator by x:
[tex]\frac{3- 2k+ \frac{k^2}{x}}{\sqrt{1+ \frac{3}{x}}+ (1+\frac{k}{x})}[/tex].

Now, as x goes to infinity, the terms with x in the denominator go to 0 so the limit is
[tex]\frac{3}{2}[/tex].

[Guest]

[tex]f(x)=\sqrt{x^{2}+3x}[/tex]
[tex]g(x)=ax+b[/tex]
[tex]\lim_{x\to\infty}[f(x)-g(x)]=0[/tex]
[tex]\lim_{x\to\infty}[f(x)-g(x)]=\lim_{x\to\infty}[\sqrt{x^{2}+3x}-(ax+b)]=\lim_{x\to\infty}[(\sqrt{x^{2}+3x}-(ax+b))\cdot1]=\lim_{x\to\infty}[(\sqrt{x^{2}+3x}-(ax+b))\cdot \frac{\sqrt{x^{2}+3x}+(ax+b)}{\sqrt{x^{2}+3x}+(ax+b)}]=\lim_{x\to\infty}[\frac{(\sqrt{x^{2}+3x}-(ax+b))\cdot(\sqrt{x^{2}+3x}+(ax+b))}{\sqrt{x^{2}+3x}+(ax+b)}]=\lim_{x\to\infty}[\frac{\sqrt{x^{2}+3x}^{2}-(ax+b)^{2}}{\sqrt{x^{2}+3x}+(ax+b)}]=\lim_{x\to\infty}[\frac{(x^{2}+3x)-(a^{2}x^{2}+2axb+b^{2})}{\sqrt{x^{2}+3x}+(ax+b)}]=\lim_{x\to\infty}[\frac{x^{2}+3x-a^{2}x^{2}-2abx-b^{2})}{\sqrt{x^{2}+3x}+(ax+b)}]=\lim_{x\to\infty}[\frac{(1-a^{2})x^{2}+(3-2ab)x-b^{2})}{\sqrt{x^{2}+3x}+(ax+b)}]=\lim_{x\to\infty}[\frac{(1-a^{2})x^{2}}{\sqrt{x^{2}+3x}+(ax+b)}+\frac{(3-2ab)x}{\sqrt{x^{2}+3x}+(ax+b)}-\frac{b^{2}}{\sqrt{x^{2}+3x}+(ax+b)}]=\lim_{x\to\infty}\frac{(1-a^{2}) \cdot x^{2}}{\sqrt{x^{2}+3x}+(ax+b)}+\lim_{x\to\infty}\frac{(3-2ab) \cdot x}{\sqrt{x^{2}+3x}+(ax+b)}-\lim_{x\to\infty}\frac{b^{2} \cdot1}{\sqrt{x^{2}+3x}+(ax+b)}=(1-a^{2})\cdot\lim_{x\to\infty}\frac{x^{2}}{\sqrt{x^{2}+3x}+(ax+b)}+(3-2ab)\cdot\lim_{x\to\infty}\frac{x}{\sqrt{x^{2}+3x}+(ax+b)}-b^{2}\cdot\lim_{x\to\infty}\frac{1}{\sqrt{x^{2}+3x}+(ax+b)}=(1-a)(1+a)\cdot\lim_{x\to\infty}\frac{x^{2}}{\sqrt{x^{2}+3x}+(ax+b)}+(3-2ab)\cdot\lim_{x\to\infty}\frac{x}{\sqrt{x^{2}+3x}+(ax+b)}-b^{2}\cdot\lim_{x\to\infty}\frac{1}{\sqrt{x^{2}+3x}+(ax+b)}=0[/tex]
[tex]\lim_{x\to+\infty}\frac{x^{2}}{\sqrt{x^{2}+3x}+(ax+b)}=\lim_{x\to+\infty}\frac{x^{2}}{\sqrt{x^2(1+\frac{3}{x})}+(ax+b)}=\lim_{x\to+\infty}\frac{x^{2}}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}+(ax+b)}=\lim_{x\to+\infty}\frac{x^{2}}{x\sqrt{1+\frac{3}{x}}+x(a+\frac{b}{x})}=\lim_{x\to+\infty}\frac{x\cdot x}{x\cdot(\sqrt{1+\frac{3}{x}}+(a+\frac{b}{x}))}=\lim_{x\to+\infty}\frac{x}{\sqrt{1+\frac{3}{x}}+(a+\frac{b}{x})}=\lim_{x\to+\infty}\frac{x}{1+a}=\frac{1}{1+a}\cdot\lim_{x\to+\infty}x[/tex]
[tex]\lim_{x\to+\infty}\frac{x}{\sqrt{x^{2}+3x}+(ax+b)}=\lim_{x\to+\infty}\frac{x}{\sqrt{x^2(1+\frac{3}{x})}+(ax+b)}=\lim_{x\to+\infty}\frac{x}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}+(ax+b)}=\lim_{x\to+\infty}\frac{x}{x\sqrt{1+\frac{3}{x}}+x(a+\frac{b}{x})}=\lim_{x\to+\infty}\frac{x\cdot 1}{x\cdot(\sqrt{1+\frac{3}{x}}+(a+\frac{b}{x}))}=\lim_{x\to+\infty}\frac{1}{\sqrt{1+\frac{3}{x}}+(a+\frac{b}{x})}=\lim_{x\to+\infty}\frac{1}{1+a}=\frac{1}{1+a}[/tex]
[tex]\lim_{x\to+\infty}\frac{1}{\sqrt{x^{2}+3x}+(ax+b)}=\lim_{x\to+\infty}\frac{1}{\sqrt{x^2(1+\frac{3}{x})}+(ax+b)}=\lim_{x\to+\infty}\frac{1}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}+(ax+b)}=\lim_{x\to+\infty}\frac{1}{x\sqrt{1+\frac{3}{x}}+x(a+\frac{b}{x})}=\lim_{x\to+\infty}\frac{1}{x\cdot(\sqrt{1+\frac{3}{x}}+(a+\frac{b}{x}))}=\lim_{x\to+\infty}\frac{1}{x\cdot(1+a)}=\frac{1}{1+a}\cdot\lim_{x\to+\infty}\frac{1}{x}=\frac{1}{1+a}\cdot0=0[/tex]
[tex]\lim_{x \to +\infty}[f(x)-g(x)]=(1-a)(1+a)\cdot\frac{1}{1+a}\cdot\lim_{x\to+\infty}x+(3-2ab)\cdot\frac{1}{1+a}-b^{2}\cdot0=(1-a)\cdot\lim_{x\to+\infty}x+(3-2ab)\cdot\frac{1}{1+a}=0[/tex]
[tex]1-a_{1} =0[/tex]
[tex]a_{1}=1-0[/tex]
[tex]a_{1}=1[/tex]
[tex]3-2a_{1}b_{1}=0[/tex]
[tex]3-2b_{1}=0[/tex]
[tex]b_{1}=\frac{3-0}{2}[/tex]
[tex]b_{1}=\frac{3}{2}[/tex]
[tex]g_{1}(x)=a_{1}x+b_{1}=x+\frac{3}{2}[/tex]
[tex]\lim_{x\to+\infty}[f(x)-g_{1}(x)]=\lim_{x\to+\infty}[f(x)-x-\frac{3}{2}]=0[/tex]
[tex]\lim_{x\to-\infty}\frac{x^{2}}{\sqrt{x^{2}+3x}+(ax+b)}=\lim_{x\to-\infty}\frac{x^{2}}{\sqrt{x^2(1+\frac{3}{x})}+(ax+b)}=\lim_{x\to-\infty}\frac{x^{2}}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}+(ax+b)}=\lim_{x\to-\infty}\frac{x^{2}}{(-x)\sqrt{1+\frac{3}{x}}+x(a+\frac{b}{x})}=\lim_{x\to-\infty}\frac{(-x)\cdot (-x)}{(-x)\cdot(\sqrt{1+\frac{3}{x}}-(a+\frac{b}{x}))}=\lim_{x\to-\infty}\frac{(-x)}{\sqrt{1+\frac{3}{x}}-(a+\frac{b}{x})}=\lim_{x\to-\infty}\frac{(-x)}{1-a}=\frac{1}{1-a}\cdot\lim_{x\to-\infty}(-x)[/tex]
[tex]\lim_{x\to-\infty}\frac{x}{\sqrt{x^{2}+3x}+(ax+b)}=\lim_{x\to-\infty}\frac{x}{\sqrt{x^2(1+\frac{3}{x})}+(ax+b)}=\lim_{x\to-\infty}\frac{x}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}+(ax+b)}=\lim_{x\to-\infty}\frac{x}{(-x)\sqrt{1+\frac{3}{x}}+x(a+\frac{b}{x})}=\lim_{x\to-\infty}\frac{(-x)\cdot (-1)}{(-x)\cdot(\sqrt{1+\frac{3}{x}}-(a+\frac{b}{x}))}=\lim_{x\to-\infty}\frac{(-1)}{\sqrt{1+\frac{3}{x}}-(a+\frac{b}{x})}=\lim_{x\to-\infty}\frac{(-1)}{1-a}=\frac{(-1)}{1-a}[/tex]
[tex]\lim_{x\to-\infty}\frac{1}{\sqrt{x^{2}+3x}+(ax+b)}=\lim_{x\to-\infty}\frac{1}{\sqrt{x^2(1+\frac{3}{x})}+(ax+b)}=\lim_{x\to-\infty}\frac{1}{\sqrt{x^2}\sqrt{1+\frac{3}{x}}+(ax+b)}=\lim_{x\to-\infty}\frac{1}{(-x)\sqrt{1+\frac{3}{x}}+x(a+\frac{b}{x})}=\lim_{x\to-\infty}\frac{1}{(-x)\cdot(\sqrt{1+\frac{3}{x}}-(a+\frac{b}{x}))}=\lim_{x\to-\infty}\frac{1}{(-x)\cdot(1-a)}=\frac{1}{1-a}\cdot\lim_{x\to-\infty}\frac{1}{(-x)}=\frac{1}{1-a}\cdot0=0[/tex]
[tex]\lim_{x \to -\infty}[f(x)-g(x)]=(1-a)(1+a)\cdot\frac{1}{1-a}\cdot\lim_{x\to-\infty}(-x)+(3-2ab)\cdot\frac{(-1)}{1-a}-b^{2}\cdot0=(1+a)\cdot\lim_{x\to-\infty}(-x)+(3-2ab)\cdot\frac{(-1)}{1-a}=0[/tex]
[tex]1+a_{2} =0[/tex]
[tex]a_{2}=0-1[/tex]
[tex]a_{2}=-1[/tex]
[tex]3-2a_{2}b_{2}=0[/tex]
[tex]3+2b_{2}=0[/tex]
[tex]b_{2}=\frac{0-3}{2}[/tex]
[tex]b_{2}=\frac{-3}{2}[/tex]
[tex]g_{2}(x)=a_{2}x+b_{2}=-x-\frac{3}{2}[/tex]
[tex]\lim_{x\to-\infty}[f(x)-g_{2}(x)]=\lim_{x\to-\infty}[f(x)+x+\frac{3}{2}]=0[/tex]