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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}

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.

5xum
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1_student
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  • try testing every step with wolfram alpha and seeing where the two differ. – 5xum Nov 20 '23 at 08:21
  • You can use $\lim_{n\rightarrow\infty} (a_n b_n) = (\lim_{n\rightarrow\infty} a_n )(\lim_{n\rightarrow\infty} b_n)$ only if both limits exists and are finite. – stange Nov 20 '23 at 08:23
  • Try using the binomial theorem. The first approach gives $$2 + O((x+3)^{-1})$$ which now can be seen to have limit 2. The second gives $$-1 + 3 + O(x^{-1}),$$ where the symbol $O()$ means "terms of order". – WA Don Nov 20 '23 at 08:29
  • Both the approaches have the same mistake. A part of an expression cannot in general be replaced by its limit during the evaluation of limit of the expression. The step where you have replaced the square root thing by its limit $1$ is invalid. However by luck your first approach gives right answer. See more details at https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Nov 20 '23 at 08:50
  • It's not true, for example, that $\lim\limits_{x\to \infty}(x(1+1/x)-x)=\lim\limits_{x\to\infty}(x-x)=0$ (this might look true since $\lim\limits_{x\to\infty}(1+1/x)=1$). In fact for that case you can see that the correct limit is $1$. Both solutions you provided make the same mistake. The moral of the story is you have to be quite careful when dealing with limits and infinity. You might want to review precisely what the limit laws do and don't say. – pancini Nov 20 '23 at 10:29
  • You recieved 3 answers to your question. Is any of them what you needed? If so, consider accepting the best answer and upvoting all useful answers you got. That's how the site works. – 5xum Nov 28 '23 at 09:38

3 Answers3

4

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.

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 $\displaystyle \lim_{x\to\infty} \frac{4+\frac{13}{x}}{\sqrt{1 + \frac6x+\frac{14}{x^2}} + 1 + \frac 1x} = \frac{4}{2} = 2.$
5xum
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1

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.

heropup
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  • Good to know I didn't mess up my algebraic manipulations :) – 5xum Nov 20 '23 at 08:43
  • Thanks for your answer.

    First Question:

    But since $x \rightarrow \infty$, we multiply it by $\frac{\infty}{\infty}$ ?

    Second Question:

    In the second step of the solution, how can we cancel the terms such as $\lim _{x \rightarrow \infty}(...x^2- x^2)...$

    Is it not $\infty-\infty$ indeterminate case?

    Could you explain the reason clearly and in detail?

     – 1_student Nov 20 '23 at 09:49
  • @1_student $x^2-x^2$ is always equal to $0$. There is no indeterminate cases. For example, if I write $\lim_{x\to\infty} x^2-x^2 = \lim_{x\to\infty} 0$, all I did was I took some expression $A(B(x))$, and then found some expression $C(x)$ which is equal to $B(x)$, and then said that $A(B(x))=A(C(x))$. This is pure algebraic manipulation. – 5xum Nov 21 '23 at 05:26
  • @1_student In other words, any statement about $x^2-x^2$ is always equally true if it is made a statement about $0$, just like any statement about George Washington is equally true if it is made a statement about the first president of the USA. – 5xum Nov 21 '23 at 05:30
  • @1_student Show me, in any of the steps in my answer, where I actually performed any multiplication by $\infty/\infty$. You can't because I did no such thing. An indeterminate form, like $\infty - \infty$, is a way to describe limiting behavior. It doesn't describe the actual function. Every step in my answer is an algebraic identity that is always true so long as I did not perform a division by zero. So as long as you substitute any value of $x$ that does not make any denominator zero, the beginning expression will always equal the ending expression. Try it for yourself. – heropup Nov 21 '23 at 06:10
0

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

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

Will Jagy
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