Solving a Limit Problem: Step-by-Step Guide

[askor]

Can someone please tell me how to solve a limit problem like this?

$$\lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}}$$

This is my attempt to solve the problem:

$$\lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}}$$
$$= \lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}} × \frac{\sqrt{x^2 + x} + \sqrt{x^2 - 3x}}{\sqrt{x^2 + x} + \sqrt{x^2 - 3x}}$$
$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{(x^2 + x) - (x^2 - 3x)}$$
$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{x^2 + x - x^2 + 3x}$$
$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{4x}$$
$$= \lim_{x \to \infty} \frac{(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{x}$$
$$= \lim_{x \to \infty} \frac{\frac{1}{\sqrt{x^2}}(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{\frac{1}{\sqrt{x^2}}(x)}$$
$$= \lim_{x \to \infty} \frac{\sqrt{\frac{x^2 + x}{x^2}} + \sqrt{\frac{x^2 - 3x}{x^2}}}{\frac{x}{x}}$$
$$= \lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{3}{x}}}{1}$$
$$= \sqrt{1 + \frac{1}{\infty}} + \sqrt{1 - \frac{3}{\infty}}$$
$$= \sqrt{1 + 0} + \sqrt{1 - 0}$$
$$= \sqrt{1} + \sqrt{1}$$
$$= 1 + 1$$
$$= 2$$

Is this correct?

 

[fresh_42]

Can someone please tell me how to solve a limit problem like this?

$$\lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}}$$

This is my attempt to solve the problem:

$$\lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}}$$
$$= \lim_{x \to \infty} \frac{4}{\sqrt{x^2 + x} - \sqrt{x^2 - 3x}} × \frac{\sqrt{x^2 + x} + \sqrt{x^2 - 3x}}{\sqrt{x^2 + x} + \sqrt{x^2 - 3x}}$$
$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{(x^2 + x) - (x^2 - 3x)}$$
$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{x^2 + x - x^2 + 3x}$$
$$= \lim_{x \to \infty} \frac{4(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{4x}$$
$$= \lim_{x \to \infty} \frac{(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{x}$$
$$= \lim_{x \to \infty} \frac{\frac{1}{\sqrt{x^2}}(\sqrt{x^2 + x} + \sqrt{x^2 - 3x})}{\frac{1}{\sqrt{x^2}}(x)}$$
$$= \lim_{x \to \infty} \frac{\sqrt{\frac{x^2 + x}{x^2}} + \sqrt{\frac{x^2 - 3x}{x^2}}}{\frac{x}{x}}$$
$$= \lim_{x \to \infty} \frac{\sqrt{1 + \frac{1}{x}} + \sqrt{1 - \frac{3}{x}}}{1}$$
$$= \sqrt{1 + \frac{1}{\infty}} + \sqrt{1 - \frac{3}{\infty}}$$
$$= \sqrt{1 + 0} + \sqrt{1 - 0}$$
$$= \sqrt{1} + \sqrt{1}$$
$$= 1 + 1$$
$$= 2$$

Is this correct?

Yes.

I wouldn't have used $\infty$ like a number, since this is not only wrong, it also supports a careless way of dealing with calculations. You have been so detailed and correct, that it hurts a bit to see $1/\infty$. Delete this line and the rest is fine.

 

[askor]

I wouldn't have used $\infty$ like a number, since this is not only wrong, it also supports a careless way of dealing with calculations. You have been so detailed and correct, that it hurts a bit to see $1/\infty$. Delete this line and the rest is fine.

So what I suppose to do after deleting the line? Please show me the proper math write.

 

[fresh_42]

So what I suppose to do after deleting the line? Please show me the proper math write.

As I said: it is perfectly fine without that line, just erase it. The long way would be
$$
\lim_{n \to \infty}\dfrac{\sqrt{1+\dfrac{1}{x}}+\sqrt{1-\dfrac{3}{x}} }{1}=\dfrac{\sqrt{1+\lim_{n \to \infty}\dfrac{1}{x}}+\sqrt{1-3\cdot\lim_{n \to \infty}\dfrac{1}{x}} }{1}=\dfrac{\sqrt{1+0}+\sqrt{1-3\cdot 0}}{1}
$$
but nobody writes these steps, although it shows what is going on, and that we use the fact that the square root function is continuous.