Question

lim_{x rightarrow infty} (sqrt{x ^{2}+x+1}-x)

Answer 1

write over 1
and then rationalize the numerator

Answer 2

that is the first step

Answer 3

i can't seem to read that properly, but is there really a fraction?

Answer 4

\[\lim_{x \rightarrow \infty}\frac{\sqrt{x^2+x+1}-x}{1}\]

Answer 5

\[\lim_{x \rightarrow \infty}\frac{(x^2+x+1)-x^2}{\sqrt{x^2+x+1}+x}=\lim_{x \rightarrow \infty}\frac{x+1}{\sqrt{x^2+x+1}+x}\]
\[=\lim_{x \rightarrow \infty}\frac{\frac{x+1}{\sqrt{x^2}}}{\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2}}}\]
\[\text{ Since } x->\infty \text{ then } \sqrt{x^2}=|x|=x \]
\[\lim_{x \rightarrow \infty}\frac{{\frac{x+1}{x}}}{\frac{\sqrt{x^2+x+1}+x}{\sqrt{x^2}}}\]
\[\lim_{x \rightarrow \infty}\frac{\frac{x}{x}+\frac{1}{x}}{\frac{\sqrt{x^2+x+1}}{\sqrt{x^2}}+\frac{x}{\sqrt{x^2}}}\]
\[\lim_{x \rightarrow \infty}\frac{1+\frac{1}{x}}{\sqrt{\frac{x^2+x+1}{x^2}}+\frac{x}{x}}\]
\[\lim_{x \rightarrow \infty}\frac{1+\frac{1}{x}}{\sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}+\frac{1}{x^2}}+1}\]
\[\lim_{x \rightarrow \infty}\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+1}\]
Let me know if you need more help :)

Answer 6

@myininaya : Tysm for your awesome help :) ...

Answer 7

Np! I hope you understood all the steps. :)

Answer 8

@myininaya : I did ,ty :)