My first limit problem

$\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^2 + x} - x$

$\displaystyle \Rightarrow \lim_{x \rightarrow \infty} x\times \Big( \sqrt{ 1 + \frac{1}{x}} - 1 \Big)$

$\displaystyle \Rightarrow \lim_{x \rightarrow \infty} \frac{\bigg( \sqrt{1 + \frac{1}{x}} - 1\bigg) }{\frac{1}{x}} \Big(\frac{0}{0} form\Big)$

Applying L'Hopital's Rule :

$\displaystyle \Rightarrow \lim_{x \rightarrow \infty}\frac{ \bigg( \frac{\frac{-1}{x^2}}{2\sqrt{1 + \frac{1}{x}}} \bigg)}{\frac{-1}{x^2}}$

$\displaystyle \Rightarrow \lim_{x \rightarrow \infty} \frac{1}{2 \sqrt{1 + \frac{1}{x}}}$

$\displaystyle \Rightarrow \frac{1}{2 \sqrt{ 1 + 0}}$

$\displaystyle \Rightarrow \boxed{\displaystyle \frac{1}{2}}$

1.- $\lim_{x\to\infty} \left(\sqrt{x^2 + x} - x \right) = \lim_{x\to\infty} \left(\sqrt{x^2 + x} - x \right) \cdot \frac{\sqrt{x^2 + x} + x}{\sqrt{x^2 + x} + x} =$ $= \lim_{x\to\infty} \frac{x^2 + x - x^2}{\sqrt{x^2 + x} + x} = \lim_{x\to\infty} \frac{ x }{\sqrt{x^2 + x} + x} =$ Now, let's divide numerator and denominator by x $= \lim_{x\to\infty} \frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{2}$

I have an intriguing way of solving this which i kinda discovered. I solved it by A.M.- G.M. inequality. $x+(1/2) \geq \sqrt{x(x+1)}$

As x tends to ∞, x ≈ x+1. Hence we can use the upper limit. Hence, the term reduces to x+(1/2)-x = 1/2. ( I doubt this step too but i leave it to the masters to guide me) This solution makes me wonder how should we deal with infinities O.o but one thing is for sure, x tends to infinity but not infinity ;)

Rationalise and then take 'x' common in numerator as well as denominator

You can show this with binomial expansion for the root: sqrt(x^2+x)=x*sqrt(1+1/x) is approx. x(1+1/2x) as 1/x->0. So, you get x+1/2-x=1/2.

$\begin{aligned} \lim_{x \to \infty} \sqrt{x^2+x}-x &= \lim_{x \to \infty} \sqrt{\left (x+\frac{1}{2} \right )^2-\frac{1}{4}}-x \\ &= \lim_{x \to \infty} \sqrt{ \left (x+\frac{1}{2} \right )^2}-x \\ &= \lim_{x \to \infty} \left (x+\frac{1}{2} \right )-x \\ &= \frac{1}{2} \end{aligned}$