Question

Does anyone know how to caucluate the sqrt? its limit question

Answer 1

Evaluate the following limits
1) \[\lim_{x \rightarrow \infty}\frac{ \sqrt{10+5x^2} }{ 6+ }\]


2)a horizontal asymptote question
\[ \lim_{x \rightarrow \infty}\sqrt{x^2+12x-10}-x\]

how do I erase the sqrt? mutiply by... the same numbers?

Answer 2

Could you rewrite the first limit? It looks like you're missing something in the denominator.

Answer 3

Second limit:
\[\lim_{x\to\infty}\left(\sqrt{x^2+12x-10}-x\right)\\
\lim_{x\to\infty}\left(\sqrt{x^2}\sqrt{1+\frac{12}{x}-\frac{10}{x^2}}-x\right)\\
\lim_{x\to\infty}\left(|x|\sqrt{1+\frac{12}{x}-\frac{10}{x^2}}-x\right)\\
\text{Since $x\to\infty$, you have $|x|=x$.}\\
\lim_{x\to\infty}\left(x\sqrt{1+\frac{12}{x}-\frac{10}{x^2}}-x\right)\\
\lim_{x\to\infty}x\left(\sqrt{1+\frac{12}{x}-\frac{10}{x^2}}-1\right)=\infty\cdot(\sqrt1-1)=\infty\cdot 0\\
\lim_{x\to\infty}\frac{\sqrt{1+\frac{12}{x}-\frac{10}{x^2}}-1}{\frac{1}{x}}=\cdots\]

Answer 4

6+4x! denometor is 6+4x

Answer 5

I would do the second like this:\[\lim_{x \rightarrow \infty}(\sqrt{x^2+12x-10} -x)\cdot \frac{ \sqrt{x^2+12x-10} +x }{ \sqrt{x^2+12x-10} +x }\]\[\lim_{x \rightarrow \infty}\frac{ x^2+12x-10-x^2 }{ \sqrt{x^2+12x-10} +x }=\]\[\lim_{x \rightarrow \infty}\frac{ 12x-10 }{ \sqrt{x^2+12x-10} +x }=\]\[\lim_{x \rightarrow \infty}\frac{ 12-\frac{ 1 }{ x } }{ \sqrt{\frac{ x^2+12x-10 }{ x^2 }} +1}=\]\[\lim_{x \rightarrow \infty}\frac{ 12-\frac{ 1 }{ x } }{ \sqrt{1+\frac{ 12 }{ x }-\frac{ 10 }{ x^2 }}+1 }=\frac{ 12-0 }{ \sqrt{1+0-0} +1}=\frac{ 12 }{ 2 }=6\]

Answer 6

Oh I see. but for the seoncd line, why the numetor is still -x^2?
isnt it +x^2?

Answer 7

I like your way to solve this equation,

Answer 8

It is a standard trick: you have something of the form a-b so you multiply with the fraction (a+b)/(a+b), making the numerator a²-b².
That is also why there is -x².

Answer 9

It is a standard trick because you can get rid of the radical in a simple way!

Answer 10

Ohhh I got it thanks!
for the first one
\[\lim_{x \rightarrow \infty} \frac{ \sqrt{10+5x^2} }{ 6+4x }=\]
so far Igot..

Answer 11

\[\lim_{x \rightarrow \infty} \frac{ \sqrt{10+5x^2} }{ 6+4x }* \frac{ \sqrt{10-5x^2} }{10-5x^2}=\]

Answer 12

then

Answer 13

\[10-5x^2/6+4x(10-5x^2)=?\]

Answer 14

I think the trick doesn't work here. It works in the other one because you have something of the form ∞-∞, which you can't compute.

In this case, I would try to put 6+4x also under the radical. That can be done like this:\[\lim_{x \rightarrow \infty}\sqrt{\frac{ 10+5x^2 }{ (6+4x)^2 }}=\lim_{x \rightarrow \infty}\sqrt{\frac{ 10+5x^2 }{ 36+48x+16x^2 }}\]Now divide numerator and denominator by the "fastest" number: that is the x with the highest power, so x²:\[\lim_{x \rightarrow \infty}\sqrt{\frac{ \frac{ 10 }{ x^2 }+5 }{ \frac{ 36 }{ x^2 }+\frac{ 48 }{ x } +16}}\]

Answer 15

Because we have divided everything by x², we can see where this leads to: \[\sqrt{\frac{ 0+5 }{ 0+0+16 }}=\sqrt{\frac{ 5 }{ 16 }}=\frac{ 1 }{ 4 }\sqrt{5}\]

Answer 16

I am confused. why (6+4x)^2?

Answer 17

Because I want to put it inside the radical

Answer 18

I.e. 4=√(4²)=√16

Answer 19

Ok, so the answer is 1/4 Srt(5)?

Answer 20

Yup!

Answer 21

i entred it and it tells me wrong :/

Answer 22

It should be correct, are you sure you entered \[\sqrt5 \over 4\]
instead of the wrong variant \[1 \over 4\sqrt5\]