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$$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$$

I have tried multiplying by $\frac{1}{\sqrt{x^2+4}}$ and it's reciprocal, but I cannot seem to find the solution. L'Hospital's doesn't seem to work either, as I keep getting rational square roots.

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  • $\begingroup$ Have you tried to implement a trig substition? $\endgroup$ – nanme Jun 28 '12 at 1:59
  • $\begingroup$ No, but that is far beyond the scope of this exercise. It's a calculus I exercise I am trying to help my friend with. $\endgroup$ – Jonathan Dewein Jun 28 '12 at 2:00
  • $\begingroup$ In a case like this, you should always do a thought-experiment to see what value the formula takes when $x$ is big, like a million. Then at least you see that the limit will certainly be $1$. $\endgroup$ – Lubin Jul 4 '12 at 21:28
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Hint $$\mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt {{x^2} + 4} }}{{x + 4}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\dfrac{{\sqrt {{x^2} + 4} }}{x}}}{{\dfrac{{x + 4}}{x}}} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\sqrt {\dfrac{{{x^2} + 4}}{{{x^2}}}} }}{{1 + \dfrac{4}{x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {1 + \dfrac{4}{{{x^2}}}} }}{{1 + \dfrac{4}{x}}}$$

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    $\begingroup$ I see. Why did you choose to divide by x? $\endgroup$ – Jonathan Dewein Jun 28 '12 at 1:59
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    $\begingroup$ @JonathanDewein Note that for large $x$, $\sqrt{x^2+4}$ is almost $x$, in the sense the $+4$ will not be of importance, so that $\sqrt{x^2}=x $ (for positive $x$). So for large values of $x$, $\sqrt{x^2+4}$ is like a "degree $1$ polynomial". In general, we can solve a limit at infinity of the quotient of two polynomials of the same degree by dividing both numerator and denominator by the $x^n$ where $n$ is the degree of the polynomials. The above reasoning motivates dividing by $x^1=x$, which solves the problem. $\endgroup$ – Pedro Tamaroff♦ Jun 28 '12 at 2:07
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$${\sqrt{x^2+4}\over x+4}={x\over x+4}\sqrt{1+(4/x^2)}$$ Can you take it from there?

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Another way to look at it is to make use of the following inequality. $$x \leq \sqrt{x^2+4} \leq x+2, \,\, \forall x \in \mathbb{R}^+$$ Hence, we have that $$\dfrac{x}{x+4} \leq \dfrac{\sqrt{x^2+4}}{x+4} \leq \dfrac{x+2}{x+4}$$ Now apply the squeeze/sandwich theorem to get $$\lim_{x \to \infty}\dfrac{\sqrt{x^2+4}}{x+4} = 1$$

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$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$=$\lim\limits_{x \to \infty} \frac{\sqrt{x^2(1 + \frac{4}{x^2})}}{x+4}$=$\lim\limits_{x \to \infty} \frac{x\sqrt{1 + \frac{4}{x^2}}}{x+4}$.

When $ x\longrightarrow\propto $ $\Rightarrow$ $\frac{4}{x^2}$$\longrightarrow$ 0 .

The above integral then takes the form:

$\lim\limits_{x \to \infty} \frac{x}{x+4}$=$\lim\limits_{x \to \infty} \frac{x}{x(1+\frac{4}{x})}$.=$\lim\limits_{x \to \infty} \frac{1}{1+\frac{4}{x}}$.

When $ x\longrightarrow\propto $ $\Rightarrow$ $\frac{4}{x}$$\longrightarrow$ 0 .

Now the above integral then takes the form:

$\lim\limits_{x \to \infty} \frac{1}{1+\frac{4}{x}}$=$\lim\limits_{x \to \infty} \frac{1}{1+0}$=1.

$\lim\limits_{x \to \infty} \frac{\sqrt{x^2 + 4}}{x+4}$=1

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HINT:

$$\left(\frac{\sqrt{x^2\:+\:4}}{x+4}\right)=\left(\frac{\left|x\right|\sqrt{1\:+\:\frac{4}{x^2}}}{x+4}\right)$$ So $$\lim _{x\to +\infty }\left(\frac{\left|x\right|\sqrt{1\:+\:\frac{4}{x^2}}}{x+4}\right) = 1$$ $$\lim _{x\to -\infty }\left(\frac{\left|x\right|\sqrt{1\:+\:\frac{4}{x^2}}}{x+4}\right) = -1$$

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Hint:

In the numerator, the dominant term of the polynomial is $x^2$, of which you take the square root. In the denominator, the dominant term is $x$. Then the function behaves like $|x|/x$ at infinity.

Or

$$\lim_{x\to\infty}\frac{\sqrt{x^2+4}}{x+4}=\lim_{x\to\infty}\frac{\sqrt{(x-4)^2+4}}x=\lim_{x\to\infty}\sqrt{1-\frac{8}{x}+\frac{20}{x^2}}.$$

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