Trick to solve this limit $\lim_{x \to\infty} \sqrt{x^2+3x}-x$

Question

The question is $$\lim_{x \to\infty} \sqrt{x^2+3x}-x$$ I divided and multiplied by $x^2$ but it gave me answer $0\cdot\infty$ which is undefined I suppose. any hint !! (PS: avoid using L Hospital's Rule)

Answer 1

HINT: Multiply by

$$\frac{\sqrt{x^2+3x}+x}{\sqrt{x^2+3x}+x}\;,$$

One more step will be required, but you can probably work it out. Just in case, I’ve included it in the spoiler-protected block below.

Then divide numerator and denominator by $x$.

Answer 2

As I said this limit is equals to: $$\lim \frac{3x}{\sqrt{x^{2}+3x}+x}$$ Now you can finish by yourself.

Answer 3

Hint

Just another way (among so many). Rewrite $$\sqrt{x^2+3x}-x=x\sqrt{1+\frac 3x}-x$$ Remember that, for small $y$, $\sqrt{1+y}\sim1+\frac y2$. Replace $y$ by $\frac 3 x$ and conclude.

Answer 4

Another way:

As $x^2+3x=\dfrac{(2x+3)^2-3^2}4,$

Let $2x+3=\sqrt3\csc2y\implies y\to0^+$

$$\lim_{x\to\infty}(\sqrt{x^2+3x}-x)=\lim_{y\to0^+}\dfrac{\sqrt3\cot2y-(\sqrt3\csc2y-3)}2$$

$$=\dfrac32-\dfrac{\sqrt3}2\lim_{y\to0^+}\dfrac{1-\cos2y}{\sin2y}$$

$$\lim_{y\to0^+}\dfrac{1-\cos2y}{\sin2y}=\lim_{y\to0^+}\dfrac{2\sin^2y}{2\sin y\cos y}=\lim_{y\to0^+}\tan y=0$$