# How do you find the limit of sqrt(9x+x^2)/(x^4+7) as x approaches oo?

Jul 31, 2016

Reqd. Lim.$= 0$.

#### Explanation:

Reqd. Limit $= {\lim}_{x \rightarrow \infty} \frac{\sqrt{9 x + {x}^{2}}}{{x}^{4} + 7}$

=lim_(xrarroo)(sqrt(x^2(9/x+1)))/(x^4(1+7/x^4)

$= {\lim}_{x \rightarrow \infty} \frac{x \sqrt{\frac{9}{x} + 1}}{{x}^{4} \left(1 + \frac{7}{x} ^ 4\right)}$

$= {\lim}_{x \rightarrow \infty} \left(\frac{1}{x} ^ 3\right) \frac{\sqrt{\frac{9}{x} + 1}}{1 + \frac{7}{x} ^ 4}$

We need to recall, here, that,

as $x \rightarrow \infty , \frac{9}{x} = 9 \cdot \frac{1}{x} \rightarrow 0 , \frac{1}{x} ^ 3 \rightarrow 0 , \mathmr{and} , \frac{7}{x} ^ 4 \rightarrow 0$.

Hence, the reqd. lim.$= 0 \cdot \left(\frac{\sqrt{0 + 1}}{1 + 0}\right) = 0 \cdot 1 = 0$.

In fact, ${\lim}_{x \rightarrow \infty} \sqrt{\frac{9}{x} + 1}$

=sqrt{lim_(xrarroo)(9/x+1)

$= \sqrt{0 + 1} = 1$.

The inter-changeability of the limit & sqrt. fun is because of the continuity of the sqrt. fun.

Jul 31, 2016

The limit exists, and it is zero.

#### Explanation:

Factor out the greatest power of $x$ from both numerator and denominator:

sqrt(9x+x^2)/(x^4+7) = sqrt(x^2(9/x+1))/(x^4(1+7/x^4)

Since $\sqrt{{x}^{2}} = | x |$, we can continue with

sqrt(x^2(9/x+1))/(x^4(1+7/x^4)) = (|x|sqrt(9/x+1))/(x^4(1+7/x^4)

Since $x$ is approaching positive infinity, we have $| x | = x$. The next step is thus

$\frac{| x | \sqrt{\frac{9}{x} + 1}}{{x}^{4} \left(1 + \frac{7}{x} ^ 4\right)} = \frac{\sqrt{\frac{9}{x} + 1}}{{x}^{3} \left(1 + \frac{7}{x} ^ 4\right)}$

At this point, we're good to go: since $x$ is approaching positive infinity, every quantity like $\frac{k}{x} ^ \alpha$ vanished, with $k$ a real number and $\alpha > 0$.

Thus, the square root approaches one:

$\sqrt{\cancel{\frac{9}{x}} + 1} \setminus \to \sqrt{1} = 1$

As for the parenthesis in the denominator, with similar claims we have

$\left(1 + \cancel{\frac{7}{x} ^ 4}\right) \setminus \to 1$

Thus, the global ratio behaves like

$\frac{1}{\infty \cdot 1} \setminus \to 0$

as $x$ approaches positive infinity.