Why is my calculation of $\lim_{x\to \infty} (\sqrt{x^2+x}-x)$ wrong?

Question

Statement :

$\lim_{x\to \infty} (\sqrt{x^2+x}-x)$

Here's my solution:

$\lim_{x\to \infty} (\sqrt{x^2(1+ \frac1x )}-x)$

$\lim_{x\to \infty} (x\sqrt{(1+ \frac1x )}-x)$

As $x\to \infty$, $\sqrt{(1+ \frac1x )}\to 1$

Therfore, as $x\to \infty$, $x -x = 0$

Answer 1

The problem lies just after the line $\lim \limits_{x \to \infty} x\sqrt{1 + \frac{1}{x}} - x$. You can't just say $x - x$ goes to $0$. Yes, it does, but we don't have $x - x$. We have $$x \cdot( \text{something}) - x$$ which needs to be factored into $$x \cdot (\text{something} - 1).$$

So you should have: \begin{split} \lim \limits_{x \to \infty} x\sqrt{1 + \frac{1}{x}} - x &= \lim \limits_{x \to \infty} x \left (\sqrt{1 + \frac{1}{x}} - 1 \right )\end{split}

Hint: Try multiplying $x(\sqrt{1 + \frac{1}{x}} - 1)$ by $1$ in the form of $\frac{(\sqrt{1 + \frac{1}{x}} + 1)}{(\sqrt{1 + \frac{1}{x}} + 1)}$.

EDIT: I just want to say in more detail what you did wrong: You took the limit as $x \to \infty$ of the expression $\sqrt{1 + \frac{1}{x}}$ first and ignored the other $x$'s. You can't do that. When you send $x$ to $\infty$, you have to send all of the $x$'s to $\infty$ at the same time. You can't pick certain $x$'s to send to $\infty$ and ignore the others. That's why when you work these problems out, you have to simplify the expression without the limit first using legal algebra, and then once you get to a new form, try taking the limit.

Answer 2

Note that it is

\begin{align}\lim_{x\to\infty}(\sqrt{x^2+x}-x)&\\ &=\lim_{x\to\infty}\frac{(\sqrt{x^2+x}-x)(\sqrt{x^2+x}+x)}{\sqrt{x^2+x}+x}\\ &=\lim_{x\to\infty}\frac{x}{\sqrt{x^2+x}+x}\\ &=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac1x}+1}\\&=\frac12.\end{align}

Where are you wrong? If $\lim_{x\to\infty} f(x)=1$ then you can't conclude that $$\lim_{x\to\infty} (xf(x)-x)=0.$$ An easier example to realize this is given by $f(x)=1+\frac 1x.$ It is $$xf(x)-f(x)=1,\forall x.$$

Answer 3

Take an example:

$\lim_{x\to\infty}(1+{1\over x})\to 1$

Does this imply $\lim_{x\to\infty}x(1+{1\over x})\to x$ ?

Hint: Break and check.

Answer 4

Limits which evaluate to something like $\infty-\infty$ or $0*\infty$ are indeterminate forms, and need to be evaluated with care.

You may have not realized it, but in your computation you evaluated the limit of the sum as the sum of the limits (when you treated the left term as just $x$). This is only justified if both limits exist.

Answer 5

The other answers are correct and they give you the correct method of evaluating the limit. In this answer I will try to give some intuition as to why your method was incorrect.

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First, let $$ f(x) = \sqrt{1+\frac1x}-1. $$ Clearly, $f(x)\to0$ as $x\to\infty$.

Now let us look at your limit. You can write it as $$ \lim_{x\to\infty} x\cdot f(x). $$ Do you see the problem? Your method says that, since $f(x)\to0$, then I can replace $xf(x)$ by zero and so the limit is zero. However, $f(x)$ is multiplied by $x$, which is going to infinity!

Roughly speaking, we want to see the relative speeds of the growth of $x$ and the decay of $f(x)$. For example, if $f(x)$ decays slower than $x$ grows, then their product goes to infinity.

So how does $f(x)$ decay? Here we can apply the useful approximation$^1$ $$ |y| \ll 1 \implies \sqrt{1+y} \approx 1+\frac y2. $$ So when $x$ is very large, $1/x$ is very small, and we can approximate $f(x)$ by $$ f(x) \approx 1 + \frac{1}{2x} - 1 = \frac{1}{2x}. $$ Therefore, when $x$ is very large, $$ x\cdot f(x) \approx x\cdot\frac{1}{2x} = \frac12. $$

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$^1$This approximation is useful in this example for gaining an understanding of the behavior of your function. Do not use it to evaluate the limit; use the methods in the other answers instead.