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I'm struggling to find the solution of the following:

$$\lim_{x\rightarrow\infty}(\sqrt{x^2+4x} - x)$$

I come to the answer of $0$.

The book has an answer of $4/1$.

The book explains a part of the question briefly. Looking at the brief answer information given I then come to an answer of $2$...

What is the answer?

Juniven Acapulco
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Luke Taylor
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5 Answers5

3

Hint: $$\sqrt{x^2+4x}- x = \frac{4x}{\sqrt{x^2+4x}+x}$$

And you are right with the answer of $2$.

Macavity
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  • Why do I have to form a fraction from the given expression? – Luke Taylor Nov 16 '13 at 20:22
  • Because then you have more options to find the limit, like cancelling out the $x$ from both numerator and denominator, or use L'Hôpital's rule, etc. – Macavity Nov 16 '13 at 20:24
  • You could do without the fraction too, but then it becomes more cumbersome. e.g. expand the square root term using the general binomial theorem. – Macavity Nov 16 '13 at 20:31
2

Just to give an alternative to multiplying by the conjugate expression, note that letting $x=u-2$ gives

$$\lim_{x\to\infty}(\sqrt{x^2+4x}-x)=2+\lim_{u\to\infty}(\sqrt{u^2-4}-u)$$

Now for $u\gt0$,

$$\left(u-{2\over u}\right)^2=u^2-4+{4\over u^2}\lt u^2-4\implies u-{2\over u}\lt\sqrt{u^2-4}\lt u$$

hence

$$-{2\over u}\lt\sqrt{u^2-4}-u\lt0$$

so by the Squeeze Theorem $\lim_{u\to\infty}(\sqrt{u^2-4}-u)=0$, and thus $\lim_{x\to\infty}(\sqrt{x^2+4x}-x)=2$.

Added later: Here is another alternative. Eschewing motivation, letting $x={1\over u+u^2}$ with $u\gt0$ gives

$$\sqrt{x^2+4x}-x=\sqrt{\left(1\over u+u^2\right)^2+{4\over u+u^2}}-{1\over u+u^2}={\sqrt{1+4u+4u^2}-1\over u+u^2}={\sqrt{(1+2u)^2}-1\over u+u^2}={1+2u-1\over u+u^2}={2\over1+u}$$

and thus

$$\lim_{x\to\infty}(\sqrt{x^2+4x}-x)=\lim_{u\to0^+}{2\over1+u}=2$$

(To motivate things, first let $x=1/h$ to obtain $(\sqrt{1+4h}-1)/h$ and then let $h=u+u^2$ to get a square inside the square root.)

Barry Cipra
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2

Multiplying by the quantity $\sqrt{x^2 + 4x} + x$ on top and bottom leads to

$$\sqrt{x^2 + 4x} - x = \frac{x^2 + 4x - x^2}{\sqrt{x^2 + 4x} + x} = \frac{4x}{\sqrt{x^2 + 4x} + x}$$

Now divide each term by $x$, noting that this becomes $x^2$ under the square root:

$$\frac{4x}{\sqrt{x^2 + 4x} + x} = \frac{\frac{4x}{x}}{\sqrt{\frac{x^2 + 4x}{x^2}} + \frac{x}{x}} = \frac{4}{1 + \sqrt{1 + \frac{4}{x}}}$$

Taking $x \to \infty$, this limit is $2$. There appears to be an error in the book, since Wolfram Alpha also returns $2$.

  • Why do I have to form a fraction from the given expression? – Luke Taylor Nov 16 '13 at 20:22
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    @LukeTaylor It's a common technique used when you have square roots - it works quite well. –  Nov 16 '13 at 20:23
  • Why can't I solve just using the original expression? – Luke Taylor Nov 16 '13 at 20:24
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    @LukeTaylor Because the original expression leads to forms like $\infty - \infty$ if we try to proceed directly, which aren't very meaningful. Another reason to introduce some form of multiplication is to take advantage of the fact that $\sqrt{ab} = \sqrt{a}\sqrt{b}$; there's no analogous result for $\sqrt{a + b}$. –  Nov 16 '13 at 20:26
  • To Luke Taylor: In limit questions specially dealing with $x \to \infty$ one can't substitute $x = \infty$ as $\infty$ is a not a number. So we have use the fact $\lim_{x \to \infty}(1/x) = 0$ and try to transform the expression into terms like $1/x$ and their powers and then put them as $0$. If that does not suit you, you can put $x = 1/y$ and let $y \to 0+$. This will get rid of confusing symbol $\infty$. – Paramanand Singh Nov 17 '13 at 05:46
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Hint

$$ \lim_{x\rightarrow \infty}(\sqrt{x^2+4x}-x)\frac{\sqrt{x^2+4x}+x}{\sqrt{x^2+4x}+x}= \lim_{x\rightarrow \infty}\frac{x^2+4x-x^2}{\sqrt{x^2+4x}+x}= \lim_{x\rightarrow \infty}\frac{4x}{x\sqrt{1+\frac{4}{x}}+x}=...$$

Avitus
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0

Putting $\displaystyle\frac1x=h,$

$$\lim_{x\to\infty}\sqrt{x^2+4x} - x=\lim_{h\to0^+}\left(\sqrt{\frac{1+4h}{h^2}}-\frac1h\right)=\lim_{h\to0^+}\left(\frac{\sqrt{1+4h}-1}h\right)$$

$$=\lim_{h\to0^+}\frac{1+4h-1}{h(\sqrt{1+4h}+1)}=4\lim_{h\to0^+}\frac1{\sqrt{1+4h}+1}\text{ as }h\ne0\text{ as }h\to0^+$$

$$=\cdots$$

lab bhattacharjee
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