- #1

- 48

- 0

## Homework Statement

[tex]\lim_{x \rightarrow \infty}[/tex][tex] 2x+1-[/tex][tex]\sqrt{4x^2+5}[/tex]

## The Attempt at a Solution

i am wondering if this method that i used is correct. i get the correct answer but i ahaven't see it in the text book or on the net. am i doing something that shouldn't be done?

using the limit laws, i take out 1.

[tex]\lim_{x \rightarrow \infty}[/tex] 1 + [tex]\lim_{x \rightarrow \infty [/tex][tex]2x-\sqrt{4x^2+5}[/tex]

then i rationalize.

.

[tex]\lim_{x \rightarrow \infty[/tex] 1 + [tex]\lim_{x \rightarrow \infty} \frac{4x^2-4x^2-5}{2x+\sqrt{4x^2+5}}[/tex]

1 + [tex]\lim_{x \rightarrow \infty} [/tex][tex]\frac{-5}{2x+x\sqrt{4+\frac{5}{x^2}}}[/tex]

1 + [tex]\lim_{x \rightarrow \infty} \frac{\frac{-5}{x}}{\frac{2x}{x}+\frac{x}{x}\sqrt{4x+\frac{5}{x^2}}}[/tex]

1 + [tex]\frac{0}{4}[/tex]

1+0=1

i get the right answer when i do it for [tex]\lim_{x \rightarrow \infty [/tex][tex]\sqrt{9x^4-3x^2+1} - 3x^2+5[/tex] and these the only two i tried. is this method just lucky for these two or can i keep using it?

also. i seem to be getting the right limit of 1 when x approches postive infinity for [tex]\lim_{x \rightarrow \infty [/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex] but i get -1 when approches negative infinity and mathematica keeps telling me its +infinity. can someone do provide me with a soloution?

Last edited: