Question

Using L'Hopitals rule solve , lim x-> infinity = sqrtx^2-1/2x+1

Answer 1

Deriving top and bottom I got, (2x/2sqrtx^2-1 )/ 2 which then I simplifeid so then it became 2x/sqrtx^2-1

Answer 2

If you want the avoid using L'Hopital's rule you can always use asymptotic behavior \[ \large x^2-1 \sim x^2 \\ \large 2x+1 \sim 2x\]

Answer 3

I can't really do that ^. I have to use the rule for these set of questions

Answer 4

\[\lim_{x \rightarrow \infty} \sqrt{x^2-1} = \infty\]\[\lim_{x \rightarrow \infty} 2x+1 = \infty\]\[(2x+1)' = 2\]
So L'Hopitals rule can be applied here, which gives:
\[\lim_{x \rightarrow \infty} \frac{\sqrt{x^2-1}}{2x+1} = \lim_{x \rightarrow \infty} \frac{(\sqrt{x^2-1})'}{(2x+1)'} = \lim_{x \rightarrow \infty} \frac{1/2*2x/\sqrt{x^2-1}}{2} = \lim_{x \rightarrow \infty} \frac{x}{2\sqrt{x^2-1}}\]also note that:
\[\lim_{x \rightarrow \infty}\sqrt{x^2-1} = x\]
Therefore the solution S is given by:
\[S = \lim_{x \rightarrow \infty}\frac{x}{2\sqrt{x^2-1}} = \lim_{x \rightarrow \infty}\frac{x}{2x} = \lim_{x \rightarrow \infty}\frac{1}{2} = \frac{1}{2}\]

Answer 5

I left it more as a side note for the ones interested since people are already replying to this answer :)

Answer 6

Question, how does sqrt x^2-1 = x?

Answer 7

\(\Large \displaystyle \lim_{x \to + \infty} \sqrt{x^2-1}= \lim x_{x \to \infty} x\sqrt{1+ \frac{1}{x^2}}=x \)

Answer 8

ah... hm...

Answer 9

Wish I understood haha... Thanks for the help though guys..

Answer 10

I also forgot to mention that when you found out the derivatives for numerator and denominator you simplified it to x / 2sqrt x^2-1. I thought it was 2x on top?

Answer 11

@Spacelimbus was basically doing the right thing. Consider
\[\lim_{x \rightarrow \infty} \sqrt{x^2 - 1}\]
This can be rewritten as:
\[\lim_{x \rightarrow \infty} x\sqrt{1-\frac{1}{x^2}} = (\lim_{x \rightarrow \infty}x)(\lim_{x \rightarrow \infty} \sqrt{1-\frac{1}{x^2}})\]
Clearly as x becomes very large, 1/x^2 becomes very small, thus the limit in the second bracket simplifies to sqrt(1) = 1. This gives
\[\lim_{x \rightarrow \infty} x\sqrt{1-\frac{1}{x^2}} = \lim_{x \rightarrow \infty}x\]
as desired.

Answer 12

The 2x is on the top, however it is multiplied by 1/2 (from the chain rule as we have a square root) so these two cancel, and the whole thing is divided by 2 (g'(x) from L'Hopital's rule).