Evaluating $\lim_{x\to+\infty} \frac{x}{\sqrt{x+1}}$ Evaluate $$\lim_{x\to+\infty} \frac{x}{\sqrt{x+1}}$$
My attempt: $$\lim_{x\to+\infty} \frac{x}{\sqrt{x+1}}=\lim_{x\to+\infty} \frac{x}{\sqrt{x}}=\lim_{x\to+\infty} \sqrt{x}=+\infty$$
Is this correct?
 A: Divide numerator and denominator by $x$ as follows
$$\lim_{x\to+\infty} \frac{x}{\sqrt{x+1}}=\lim_{x\to+\infty} \frac{\dfrac xx}{\dfrac{\sqrt{x+1}}{x}}=\lim_{x\to+\infty} \frac{1}{\sqrt{\frac1x+\frac1{x^2}}}=\infty$$
Alternatively, let $x=\frac1t$
$$\lim_{x\to+\infty} \frac{x}{\sqrt{x+1}}=\lim_{t\to 0}\frac{\frac1t}{\sqrt{\frac1t+1}}=\lim_{t\to 0} \frac{1}{\sqrt{t^2+t}}=\infty$$
A: The function $\sqrt{\cdot}: \mathbb{R}_{\geq 0} \to \mathbb{R}$ is increasing, and so $x/\sqrt{x+1} \geq x/\sqrt{2x}$ (when $x \geq 1$, though your limit is to infinity so it suffices to look at the function on the interval $[1,+\infty)$). Then, $x/\sqrt{2x} = \sqrt{x}/\sqrt{2} \to \infty$ as $x \to \infty$. Since $x/\sqrt{x+1} \geq x/\sqrt{2x}$, we get that the evaluation of your limit is $+\infty$.
A: You can't just delete de 1. A way is by the L'Hopital, it is deriving numerator and denominator
$\lim_{x\to \infty}\frac{x}{\sqrt{x+1}}=\lim_{x\to \infty}\frac{1}{\frac{1}{2\sqrt{x+1}}}=\lim_{x\to \infty}2\sqrt{x+1}=\infty$
A: Set $y^2=x+1,$ and consider $y \rightarrow \infty$.
$F(y):=\frac{y^2-1}{y}=$
$\frac{(y-1)(y+1)}{y}> \frac{(y-1)y}{y}=y-1.$
Take the limit.
