2
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ 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
6
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ I see. Why did you choose to divide by x? $\endgroup$ – Jonathan Dewein Jun 28 '12 at 1:59
  • 1
    $\begingroup$ @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
4
$\begingroup$

$${\sqrt{x^2+4}\over x+4}={x\over x+4}\sqrt{1+(4/x^2)}$$ Can you take it from there?

$\endgroup$
4
$\begingroup$

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 $$\lim_{x \to \infty}\dfrac{\sqrt{x^2+4}}{x+4} = 1$$

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.