Question

Evaluate the limit:

Answer 1

\[\lim_{x \rightarrow+ \infty} = \frac{ 2x + 1 }{ \sqrt{x ^{2}+3} }\]

Answer 2

This one needs a little manipulation to make it work.\[\lim_{x \rightarrow \infty}\frac{2x+1}{\sqrt{x^2+3}}=\lim_{x \rightarrow \infty}\frac{2x+1}{x \sqrt{1+\frac{3}{x^2}}}=\lim_{x \rightarrow \infty}\frac{2+\frac{1}{x}}{\sqrt{1+\frac{3}{x^2}}}\]You should be able to handle it from there.

Answer 3

\[\lim_{x \rightarrow+ \infty} = \frac{ 2+ (+\infty) }{ \sqrt{1+(+\infty)} } = \infty \]
right?

@AnimalAin

Answer 4

u need to know one thing that
\[\lim_{x \rightarrow \infty}=\lim_{(1/x) \rightarrow 0}\]

Answer 5

Not correct. Note that you are taking the limit as x approaches infinity. The reciprocal of a number going to infinity approaches zero. The answer is two, as noted earlier.

Answer 6

\[\lim_{x\rightarrow\infty} \dfrac{2x+1}{\sqrt{x^2+3}} = \lim_{x\rightarrow\infty}\dfrac{2x}{\sqrt{x^2}}\]Because when x approaches to infinity, terms with lower degree become less and less significant

Answer 7

Ok now how it will compensate \[\infty \] in the equation?

Answer 8

If you look at my work earlier, we divided out any value of x that wasn't in a denominator. As x goes to infinity, any bounded number divided by x (or x^2) will go to zero. After we get rid of those terms, we end up with 2/1 = 2.

Answer 9

aha .. so the solution = 2 ?

Answer 10

Can you give me a solution in a simplified way?

Answer 11

I thought I did. You need to look carefully at the algebra I used to manipulate the problem in a way that made the values that go to infinity go away. Keep after it; it's not that hard. I mean, if I can do it, how hard could it be? Do math every day.

Answer 12

In any case, thank you very much :*

Answer 13

No sweat.

Answer 14

Well, I kind of did. For limit as x approaches to infinity, you just need to look and compare degree (highest power) of numerator and denominator.

If degree in numerator is higher than denominator's, then limit is infinity.

If degree in denominator is higher than numerator's, then limit is 0

If both have same degree, then limit is ratio of coefficient of leading term

In this problem, both have same degree, so limit is 2/1 = 2

Answer 15

Can we do that in the equation?
\[\lim_{ء \rightarrow +\infty} = \frac{ 2x+1 }{ (x ^{2}+3)^{1/2} }\]

Answer 16

thats just the same question

Answer 17

Yes. You just focus on leading term in both numerator and denominator.
you can see that \(\sqrt{x^2}\) = x, ratio of coefficient is 2/1 = 1.

Does that make sense? Sorry if my English is bad lol

Answer 18

so
\[\lim_{x \rightarrow +\infty} \frac{ 2x + 1 }{ \sqrt{x ^{2} + 3} } = \lim_{x \rightarrow +\infty} \frac{ 2x+1 }{ \sqrt[x]{1+\frac{ 3 }{ x ^{2} }} }\]

Answer 19

\[\lim_{x \rightarrow +\infty} \frac{ 2+\frac{ 1 }{ x } }{ \sqrt{1+\frac{ 3 }{ x ^{2}}} }\]

Answer 20

\[\Large \lim_{x\rightarrow\infty}\dfrac{2x+1}{\sqrt{x^2+3}} = \lim_{x\rightarrow\infty}\dfrac{2x}{\sqrt{x^2}} = \lim_{x\rightarrow\infty}\dfrac{2x}{x} = \lim_{x\rightarrow\infty} 2 = \boxed{2}\]

Answer 21

thats it.... now when you substitute

\[\frac{1}{x} .. and..\frac{3}{x^2}\]

both approach zero as x approaches infinity

Answer 22

Guess my approach isn't clear

Answer 23

so a simple method... was to factor the term inside the radical



\[x^2 + 3 = x^2(1 + \frac{3}{x^2})\]

then applying a radical law to the factored form you get

\[x \sqrt{1 + \frac{3}{x^2}}\]

now divide every term by x

and you get

\[\frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x^2}}}\]

so now in this form... as x approaches infinity... the value of the fractions approaches zero

Answer 24

which means you limit becomes \[\frac{2}{1}\]

Answer 25

yeh

Answer 26

Thank you all :*