Question

solve the limit please
file attached

Answer 1

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

Answer 2

@phi

Answer 3

you can try multiply and divide top by
\[ \sqrt{x^2+1}+x\]

Answer 4

how?

Answer 5

multiply your expression by
\[ \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\]

are you asking what is
\[ (\sqrt{x^2-1} -x)\cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}\]?

Answer 6

*
\[ (\sqrt{x^2+1} -x)\cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x} \]

Answer 7

if so, use
(a-b)(a+b) = a^2 - b^2

Answer 8

\[\lim_{x \rightarrow \infty}=\frac{ 1 }{\sqrt{x ^{2}+1}+x }= 0\]

Answer 9

yes