Answer:
[tex]L=\frac{1}{\sqrt{x+1}+\sqrt{x}}[/tex]
No, It is not true because when x tends to infinity then L tends to 0.
Step-by-step explanation:
We are given that
[tex]A(x,\sqrt{x}[/tex]) and B[tex](x,\sqrt{x+1}[/tex])
We have to express distance L between the points A and B as a function of x
and we have to find L tends to infinity when x tends to infinity.
Distance formula between two points is given by
[tex]=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using distance formula
[tex]L=\sqrt{(x-x)^2+(\sqrt{x+1}-\sqrt{x})^2}[/tex]
[tex]L=\sqrt{(\sqrt{x+1}-\sqrt{x})^2}=\sqrt{x+1}-\sqrt{x}[/tex]
[tex]L=\frac{(\sqrt{x+1}-\sqrt{x})(\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}[/tex]
Using rationalization
Now, we get
[tex]L=\frac{(\sqrt{x+1})^2-(\sqrt{x})^2}{\sqrt{x+1}+\sqrt{x}}[/tex]
Using identity
[tex](a+b)(a-b)=a^2-b^2[/tex]
[tex]L=\frac{x+1-x}{\sqrt{x+1}+\sqrt{x}}[/tex]
[tex]L=\frac{1}{\sqrt{x+1}+\sqrt{x}}[/tex]
[tex]\lim_{x\rightarrow\infty}L=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}[/tex]
[tex]\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x}(\sqrt{1+\frac{1}{x}}+1)}[/tex]
[tex]=0[/tex]
Therefore,when x tends to infinity then L does not tends to infinity.
