$\infty-\infty$ indeterminate case

Question

The methods yield two different answers. Could you explain the reason clearly and in detailed?

Question:

$$\lim_{{x \to \infty}} \left( \sqrt{x^2 + 6x + 14} - (x+1) \right) = ?$$

Solution:

Method 1:

By employing the conjugate of the original problem, the solution is as follows:

\begin{aligned} & =\operatorname{lim}_{x \rightarrow \infty}\left[\sqrt{(x+3)^2+5}-(x+1)\right] \\ & =\lim _{x \rightarrow \infty}\left[\sqrt{(x+3)^2 \cdot\left(1+\frac{5}{(x+3)^2}\right)}-(x+1)\right] \\ & =\lim _{x \rightarrow \infty}\left[|x+3| \cdot \sqrt{1+\frac{5}{(x + 3)^2}}-(x+1)\right] \\ & =\lim _{x \rightarrow \infty}[(x+3)-(x+1)]=2 \end{aligned}

Method 2:

Alternatively,

\begin{aligned} & =\lim_{x \rightarrow \infty}\left[\sqrt{x^2 \cdot\left(1+\frac{6}{x}+\frac{14}{x^2}\right)}-(x+1)\right] \\ & =\lim_{x \rightarrow \infty} \left[|x| \cdot \sqrt{1+\frac{6}{x}+\frac{14 }{x^2}}-(x+1)\right] . \\ & =\lim_{x \rightarrow \infty}[(x)-(x+1)]=-1 \end{aligned}

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In some books, the authors solve the examples with $\infty-\infty$ indeterminate case by multiplying and dividing the expression by its conjugate in the $\infty-\infty$ indeterminate case. But is it not $\frac{\infty}{\infty}$? How can we do it? Please see my other question in this site.

Answer 1

In your first method, you replace $$(x+3)\sqrt{1+\frac{5}{(x+3)^2}}$$ with just $x+3$ without justification.

You used an "invalid" rule where you only calculate the limit of part of your expression, i.e. you replace $$\lim f(g(x)\cdot h(x))$$ with $$\lim f([\lim g(x)]\cdot h(x))$$ but there is no rule that allows you to do this. There are many limit laws, but you must be careful to always know which one you are applying.

In the second approach, you do the same just with a different set of functions.

So both your methods are incorrect, even though the first gives, by chance, the correct result.

Remember, limit laws such as $\lim a_n b_n=\lim a_n \lim b_n$ only hold if the individual limits exist! In your case, the limit of $(x+3)$ does not exist, so you cannot just replace the expressions willy nilly.

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For a better way of solving the problem, you can multiply your expression with $\frac{\sqrt{x^2+6x+1} + (x+1)}{\sqrt{x^2+6x+1} + (x+1)}$ and you should get, after some simplification, that the original expression is equal to $$\frac{4+\frac{13}{x}}{1+\frac1x + \sqrt{1+\frac6x+\frac{14}{x^2}}}$$

and this is an expression where you can use all the limit laws you know because all limits exist and are finite. In other words, once you did this simplification, you can note that

1. $\displaystyle\lim_{x\to\infty} 1 + \frac6x+\frac{14}{x^2} = 1$
2. Therefore, because $\sqrt{}$ is a continuous function around $1$, $\displaystyle\lim_{x\to\infty} \sqrt{1 + \frac6x+\frac{14}{x^2}} = \sqrt{1}=1$
3. $\displaystyle\lim_{x\to\infty} 1 + \frac1x = 1$
4. From 2 and 3, because both limits exist, we have $\displaystyle \lim_{x\to\infty} \sqrt{1 + \frac6x+\frac{14}{x^2}} + 1 + \frac 1x = 1 + 1 = 2$
5. $\displaystyle \lim_{x\to\infty} 4+\frac{13}{x} = 4$
6. From 4 and 5, and because both limits exist and the denominator's limit is not $0$, we finally conclude that $\displaystyle \lim_{x\to\infty} \frac{4+\frac{13}{x}}{\sqrt{1 + \frac6x+\frac{14}{x^2}} + 1 + \frac 1x} = \frac{4}{2} = 2.$

Answer 2

Neither method is correct for the reasons already described. You cannot write $$\lim_{x \to \infty} a(x) b(x) = \lim_{x \to \infty} a(x) \lim_{x \to \infty} b(x)$$ unless both of the right-hand limits exist.

A correct approach is to multiply the indeterminate form by its conjugate: $$\begin{align} \sqrt{x^2 + 6x + 14} - (x+1) &= \frac{\left(\sqrt{x^2 + 6x + 14} - (x+1)\right)\left(\sqrt{x^2 + 6x + 14} + (x+1)\right)}{\sqrt{x^2 + 6x + 14} + (x+1)} \\ &= \frac{x^2 + 6x + 14 - (x+1)^2}{\sqrt{x^2 + 6x + 14} + (x+1)} \\ &= \frac{4x + 13}{\sqrt{x^2 + 6x + 14} + (x+1)} \\ &= \frac{4 + \frac{13}{x}}{\sqrt{1 + \frac{6}{x} + \frac{14}{x^2}} + 1 + \frac{1}{x}}. \end{align}$$ And now we can take the limit without issues.

Answer 3

for $x > 0$ $$ \left( x + 3 \right)^2 < x^2 + 6x + 14 < \left( x + 3 +\frac{5}{2x} \right)^2 $$ so

$$ x + 3 \; \; < \; \; \sqrt{x^2 + 6x + 14} \; \; < \; \; x + 3 +\frac{5}{2x} $$