Computing Limits at Infinity Part IV

[Hckyplayer8]

Compute the lim as x --> infinity of sqrt(x²+5x) -2

Based on the example problems, the first step is to multiply by the conjugate. This leads to the lim as x--> infinity of [(x²+5x) -4] / sqrt(x² + 5x) +2

The example diverges from this practice because the numerator worked out to 1 and the denominator still had the variable so the answer ended up being zero. I don't see anything that simple with this problem.

With the square root of the variable in the denominator, that means the numerator will always be larger than the denominator which means this limit should be positive infinity which means the limit does not exist.

But, how can I prove it Algebraically like my instructor would want to see on the test?

 

[Harry_the_cat]

Is there meant to be a division sign in your original limit?

The technique of multiplying by the conjugate is really only useful if the expression is in the denominator.

There's no need to do it here.

As x approaches infinity, x²+5x approaches infinity, so therefore sqrt(x²+5x) approaches infinity. The -2 is irrelevant in the big scheme of things. So as x approaches infinity so does the expression and so the limit doesn't exist. The expression just keeps getting larger and larger.

 

[Hckyplayer8]

Is there meant to be a division sign in your original limit?

The technique of multiplying by the conjugate is really only useful if the expression is in the denominator.

There's no need to do it here.

As x approaches infinity, x²+5x approaches infinity, so therefore sqrt(x²+5x) approaches infinity. The -2 is irrelevant in the big scheme of things. So as x approaches infinity so does the expression and so the limit doesn't exist. The expression just keeps getting larger and larger.

No division sign.

Thanks!