Solve Easy Limit Problem: \sqrt{5} Neg.

[DieCommie]

$$\lim_{x\rightarrow -\infty\\} \frac{\sqrt{5x^2-2}}{x+3}$$

When I first looked at it i thought the top will increase to inf. at a rate of $$\sqrt5$$ and the bottom will increase to inf. at a rate of -1, thus the answer would be $$-\sqrt5$$

However when i do the algebra i get this...
$$\lim_{x\rightarrow -\infty\\} \frac{\sqrt{5x^2-2}}{x+3}$$

$$\lim_{x\rightarrow -\infty\\} \frac{x\sqrt{5-2/x^2}}{x(1+3/x)}$$

$$\frac{\sqrt{5-0}}{1+0}=\sqrt5$$ but the answer is negative

 

[Curious3141]

A subtle error to be sure. Because, by convention, the result of the square root operation on real positive numbers is always positive, note that $$\sqrt{x^2} = x$$ only holds for positive real x. For negative real x, $$\sqrt{x^2} = -x$$

In general, for real x, $$\sqrt{x^2} = |x|$$. This is simply because of the convention by which the square root is defined.

Can you now see your error ?

 

[DieCommie]

The x I factored out on the top, was a negative x?

 

[robert Ihnot]

As pointed out at http://mathworld.wolfram.com/LHospitalsRule.html, this is one of those rare cases where L'Hospital's rule fails. The example given there as u goes to infinity is:
$$\frac{u}{\sqrt{u^2+1}}$$

where the limit is 1. Well, since u can increase beyond bound, it is hard to see how the limit would be different if the case was: $$\frac{u+3}{\sqrt{u^2+15}}$$, which suggests a start on the problem. However I see that you have the right idea! but, since a square is positive, we have a negative value in the numerator of the original problem as pointed out by Curious 341.

 

[Curious3141]

The x I factored out on the top, was a negative x?

Yes, it should be replaced by $-x$.

 

[DieCommie]

Thx for the help

I don't know what L'Hospital's rule is, I may learn that though

 

[Curious3141]

Thx for the help

I don't know what L'Hospital's rule is, I may learn that though

It's not relevant to this question, but basically it enables you to sort out limits of quotients of functions that look indeterminate from mere inspection. The limit you posted is of a quotient, but it's pretty trivial to solve with mere inspection after a little algebra.

Certain conditions have to be met by the functions before LH rule can be invoked, and LH rule doesn't always help. It never really *fails* (as in gives a wrong answer) though, as long as the conditions are met.

Read more about it here.