Limit Infinity Square Root

[Jason76]

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(9x^{6} - x)^{1/2}}{x^{3} + 5}\)

I know that getting a variable to any power under a number (in a small fraction within the big one) will produce zero, in this situation (after infinity is plugged in).

 

[daon2]

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(9x^{6} - x)^{1/2}}{x^{3} + 5}\)

I know that getting a variable to any power under a number (in a small fraction within the big one) will produce zero, in this situation (after infinity is plugged in).

Divide the numerator and denominator by \(\displaystyle x^3\). For the numerator, use the fact that for positive x, \(\displaystyle x^3=\sqrt{x^6}\).

 

[Jason76]

Divide the numerator and denominator by \(\displaystyle x^3\). For the numerator, use the fact that for positive x, \(\displaystyle x^3=\sqrt{x^6}\).

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(9x^{6})^{1/2} - (x)^{1/2}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9x^{3} - (x)^{1/2}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9x^{3}/ x^{3} - (x)^{1/2}/ x^{3}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9x^{3}/ x^{3} - 1/x^{2}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9 - 0}{1 + 0} = 9\) - Answer




This looks like the right direction.

 

[Deleted member 4993]

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(9x^{6})^{1/2} - (x)^{1/2}}{x^{3} + 5}\)........................................Incorrect (bad algebra)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9x^{3} - (x)^{1/2}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9x^{3}/ x^{3} - (x)^{1/2}/ x^{3}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9x^{3}/ x^{3} - 1/x^{2}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{9 - 0}{1 + 0} = 9\) - Answer




This looks like the right direction.

.

 

[Jason76]

.

How about?

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(9x^{6})^{1/2} - (x)^{1/2}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{3x^{3} - (x)^{1/2}}{x^{3} + 5}\) - correction on this line. It's NOT \(\displaystyle 9x^{3}\), but rather, \(\displaystyle 3x^{3}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{3x^{3}/ x^{3} - (x)^{1/2}/ x^{3}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{3x^{3}/ x^{3} - 1/x^{2}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{3 - 0}{1 + 0} = 3\) - Answer

 

[srmichael]

How about?

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(9x^{6})^{1/2} - (x)^{1/2}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{3x^{3} - (x)^{1/2}}{x^{3} + 5}\) - correction on this line. It's NOT \(\displaystyle 9x^{3}\), but rather, \(\displaystyle 3x^{3}\) NO!!!!

\(\displaystyle \lim x \rightarrow \infty \dfrac{3x^{3}/ x^{3} - (x)^{1/2}/ x^{3}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{3x^{3}/ x^{3} - 1/x^{2}}{x^{3}/ x^{3} + 5/ x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{3 - 0}{1 + 0} = 3\) - Answer

\(\displaystyle \sqrt{a^2-b^2} \ne a-b\)

 

[Deleted member 4993]

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{(9x^{6} - x)^{1/2}}{x^{3} + 5}\)

I know that getting a variable to any power under a number (in a small fraction within the big one) will produce zero, in this situation (after infinity is plugged in).

\(\displaystyle \displaystyle \lim_{x \to \infty}\frac{\sqrt{9x^6 -x}}{x^3 + 5}\)

\(\displaystyle = \ \displaystyle \lim_{x \to \infty}\frac{\sqrt{x^6*(9 -\frac{1}{x^5})}}{x^3 *(1 + \frac{5}{x^3})}\)

\(\displaystyle = \ \displaystyle \lim_{x \to \infty}\frac{x^3*\sqrt{9 -\frac{1}{x^5}}}{x^3 *(1 + \frac{5}{x^3})}\)

\(\displaystyle = \ \displaystyle \lim_{x \to \infty}\frac{\sqrt{9 -\frac{1}{x^5}}}{1 + \frac{5}{x^3}}\)

\(\displaystyle = \ \displaystyle \frac{\sqrt{9 - 0}}{1 + 0}\)

Now finish it.....

 

[Jason76]

Taking the square root to the 1/2 power didn't work due to the + and - symbols.

 

[Jason76]

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6}/x^{3} - x/x^{3}}}{x^{3}/x^{3} + 5/x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{3} - 1/x^{2}}}{x^{3}/x^{3} + 5/x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{3} - 0}}{1 + 0}\)

 

[HallsofIvy]

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6} - x}}{x^{3} + 5}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{6}/x^{3} - x/x^{3}}}{x^{3}/x^{3} + 5/x^{3}}\)

No, this is NOT what Subhotosh Khan told you to do! You have to divide numerator and denominator by the same thing. Here you have divided the denominator by \(\displaystyle x^3\) and the numerator by \(\displaystyle \sqrt{x^3}= x^{3}{2}\). Divide both numerator and denominator by \(\displaystyle x^3\). In the numerator that goes into the square root and becomes "x^6":
\(\displaystyle \frac{\frac{1}{x^3}\sqrt{9x^6- x}}{\frac{1}{x^3}(x^3+ 5)}= \frac{\sqrt{\frac{1}{x^6}(9x^6- x)}{x^3}(x^3+ 5)}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{3} - 1/x^{2}}}{x^{3}/x^{3} + 5/x^{3}}\)

\(\displaystyle \lim x \rightarrow \infty \dfrac{\sqrt{9x^{3} - 0}}{1 + 0}\)