What's wrong with this limit solution?

Question

I was solving this limit: $$ \lim_{x\rightarrow +\infty} \left( \sqrt{x^2+x+1}-x \right) $$ And here is my solution: $$ \lim_{x\rightarrow +\infty} \left( \sqrt{x^2+x+1}-x \right) =\lim_{x\rightarrow +\infty} \left[ \sqrt{x^2\left( 1+\frac{1}{x}+\frac{1}{x^2} \right)}-x \right] =\lim_{x\rightarrow +\infty} \left( \sqrt{x^2}-x \right) =0 $$ The correct answer is 1/2.

I know the correct solution to get the right answer. I just want to know where this solution went wrong.

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Thanks for the responses. After reading some answers, I have developed some ideas, and i don't know if they are right...

When $ x\rightarrow \infty $, I always choose to neglect terms with lower orders than x. For example,$$ \lim_{x\rightarrow \infty} \frac{\left( 1+2x \right) ^{10}\left( 1+3x \right) ^{20}}{\left( 1+6x \right) ^{15}} $$ I would transform it into$$ \lim_{x\rightarrow \infty} \frac{\left( 2x \right) ^{10}\left( 3x \right) ^{20}}{\left( 6x \right) ^{15}} $$ And the calculation result is correct. If you want to write down each step in detail, it should be like this：$$ \lim_{x\rightarrow \infty} \frac{\left( 1+2x \right) ^{10}\left( 1+3x \right) ^{20}}{\left( 1+6x \right) ^{15}}=\frac{\lim_{x\rightarrow \infty} \left( 1+2x \right) ^{10}\lim_{x\rightarrow \infty} \left( 1+3x \right) ^{20}}{\lim_{x\rightarrow \infty} \left( 1+6x \right) ^{15}}=\frac{\lim_{x\rightarrow \infty} \left( 2x \right) ^{10}\lim_{x\rightarrow \infty} \left( 3x \right) ^{20}}{\lim_{x\rightarrow \infty} \left( 6x \right) ^{15}}=\lim_{x\rightarrow \infty} \frac{\left( 2x \right) ^{10}\left( 3x \right) ^{20}}{\left( 6x \right) ^{15}} $$ But the condition for me to be able to do this is that every limit here exists. Returning to the original question, if I write out the calculation process in detail, it would look like this$$ \lim_{x\rightarrow +\infty} \left( \sqrt{x^2+x+1}-x \right) =\lim_{x\rightarrow +\infty} \left[ \sqrt{x^2\left( 1+\frac{1}{x}+\frac{1}{x^2} \right)}-x \right] =\lim_{x\rightarrow \infty} \sqrt{x^2}\cdot \lim_{x\rightarrow \infty} \sqrt{\left( 1+\frac{1}{x}+\frac{1}{x^2} \right)}-\lim_{x\rightarrow \infty} x $$ However, $ \lim_{x\rightarrow \infty} \sqrt{x^2} $ and $ \lim_{x\rightarrow \infty} x $ do not exist, so the second equal sign is not valid.

I'm not sure if this is correct...

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I think I can expand on this question.

The reason I calculate this way is because I'm accustomed to discarding lower-order terms, and this approach has consistently proven effective in previous calculations.

Like $ \lim_{x\rightarrow \infty} \frac{4x^2-1}{2x^2-x-1}=\lim_{x\rightarrow \infty} \frac{4x^2}{2x^2}=\frac{1}{2} $ and $ \lim_{x\rightarrow \infty} \frac{\left( 2-3x \right) ^3\left( 3+2x \right) ^5}{\left( 1-6x \right) ^8}=\lim_{x\rightarrow \infty} \frac{\left( -3x \right) ^3\left( 2x \right) ^5}{\left( -6x \right) ^8}=-\frac{1}{2^3\times 3^5} $, their calculation results are all correct.

However, in this particular question, this method is not applicable.

So, my question is whether this method is actually correct? If it's not correct, why did it work in the previous equations? If it is correct, then why did it fail in this particular question?

Answer 1

The problem with your solution is $$\lim_{x\to+\infty}\left[\sqrt{x^2\color{red}{\left(1+\dfrac1x+\dfrac1{x^2}\right)}}-x\right]=\lim_{x\rightarrow +\infty} \left( \sqrt{x^2\color{red}{(1)}}-x \right)$$ You just evaluate the limit of a "part of the function" while neglecting other terms.

Let me give you an illustrative example. Following your argument we could get $$\lim_{x\to 0}\dfrac{x}{x}=\lim_{x\to 0}\dfrac{\color{red}{0}}{x}=0$$ which is not correct.

Answer 2

You have a problem because you cannot neglect the terms under the radical as both terms in the difference have same magnitude order. But you can use the good old equality: $a - b = \frac{a^2 - b^2}{a+b}$ (assuming $a+b \neq 0$).

Here you get:

$$\sqrt{x^2 + x + 1} - x = \frac{(x^2+x+1) -x^2}{\sqrt{x^2 + x + 1} + x} = \frac{x+1}{\sqrt{x^2 + x + 1} + x}$$

You can now safely take the limit:

$$\lim_{x\rightarrow +\infty} \left( \sqrt{x^2+x+1}-x \right) = \lim_{x\rightarrow +\infty}\frac{x+1}{\sqrt{x^2 + x + 1} + x} = \lim_{x\rightarrow +\infty}\frac{x}{x+x} = \frac 1 2 $$

Answer 3

When you took the term {x^2(1+1/x+1/x^2)}^(1/2) and approximated it as root of x^2, the x which tends to infinity here multiplied with the 1/x term which would tend to 0 inside the root thereby giving an error in the answer...

I would recommend to consider all the terms and then form a perfect square since you can neglect a constant with respect to an infinity inside the root in the following manner:

1. Instead of taking the x common, make a perfect square, in this case being [(x+1/2)^2 + 3/4].

2)Now you can take the entire squared term common and consider the entire term as (x+1/2) and since its a root of 1 and with respect to that the other term being very small, the entire root term converges to 1

3)(x+1/2) - x = 1/2

Answer 4

If you write your expression in the form $$\lim_{x\to\infty} x\left(\sqrt{1+\frac1x+\frac1{x^2}}-1\right)$$ you can see $\infty\cdot0$ indeterminacy more clearly. You can not cancel fractional terms. Multiplication by conjugate inside paranthesis results in the expression $$\lim_{x\to\infty} x\frac{\frac1x+\frac1{x^2}}{\sqrt{1+\frac1x+\frac1{x^2}}+1}$$ from which we can find the limit after the multiplication in the numerator.