Question

Compute the limit of the following as x approaches infinity.

After multiplying both sides by the conjugate, I got the limit 1/∞ after plugging in the limit, but is that the proper answer? 1/∞ would just make the answer 0, right?

Answer 1

Yes, that's the correct limit.

`√(x^2+1)-x=\frac1{√(x^2+1)+x}=\frac1{√(x^2)\sqrt{1+\frac1{x^2}}+x}`

`√(x^2)=|x|`, but since
`x\to\infty`, we are considering
`x>0`, for which
`|x|=x`. Then

`\displaystyle\lim_(x\to\infty)√(x^2+1)-x=\lim_(x\to\infty)\frac1{x\left(\sqrt{1+\frac1{x^2}}+1\right)}`

and both
`\frac1x` and
`\frac1{x^2}` vanish as
`x\to\infty`, making the overall limit 0.