Question

lim x-> infinity ((\sqrt(x^2-16))-(\sqrt(x^2-1)))

Answer 1

\[\lim x \rightarrow \infty (\sqrt{x^2+16}-\sqrt{x^2-1})\]

Answer 2

@JimSacc HINT:- multiply and divide by x

Answer 3

let me try that i think i know what you mean

Answer 4

nah i don't know whats the next step

Answer 5

\[x*\frac{ \sqrt{x^2+16}-\sqrt{x^2-1)} }{ x }\]
x*\[x*(\sqrt{\frac{ x^2+16 }{ x^2 }} - \sqrt{\frac{ x^2-1 }{x^2 }})\]

Answer 6

\[x*(\sqrt{1+\frac{ 16 }{ x^2 }} - \sqrt{1-\frac{ 1 }{ x^2 }})\]
now apply the limits

Answer 7

yeah so when it says x goes to infinity what number do i plug for x, i apologize for my complete ignorance

Answer 8

u need to plug in infinity

Answer 9

oh alright 0... haven't been studying appropriately at all this semester

Answer 10

yeah 0 will be an answer

Answer 11

@niksva, plugging \(\infty\) in directly will yield \(0\cdot\infty\), not \(0\).

@JimSacc, multiply by conjugate/conjugate:
\[\lim_{x\to\infty}\left(\sqrt{x^2-16}-\sqrt{x^2-1}\right)\cdot\frac{\sqrt{x^2-16}+\sqrt{x^2-1}}{\sqrt{x^2-16}+\sqrt{x^2-1}}\\
\lim_{x\to\infty}\frac{\left(x^2-16\right)-\left(x^2-1\right)}{\sqrt{x^2-16}+\sqrt{x^2-1}}\\
-15\lim_{x\to\infty}\frac{1}{\sqrt{x^2-16}+\sqrt{x^2-1}}\]
Now the function properly approaches 0 as \(x\to\infty\).

Answer 12

Which do you mean:

1) lim x-> infinity ((\sqrt(x^2-16))-(\sqrt(x^2-1))) or
2)\[\lim x \rightarrow \infty (\sqrt{x^2+16}-\sqrt{x^2-1})\]

I ask this because in one expression you have Sqrt(x^2+16) and in the other you have Sqrt(x^2-16).

Answer 13

@SithsAndGiggles any number multiplies with zero is always a zero

Answer 14

That is not the case, in general. Consider
\[\lim_{x\to\infty}e^xe^{-x}\]
Plugging in directly yields \(0\cdot\infty\). However, \(e^{x}e^{-x}=1\), not \(0\).

Answer 15

If the limit is as @mathmale indicated:

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
@JimSacc, multiply by conjugate/conjugate:
\[\lim_{x\to\infty}\left(\sqrt{x^2\color{red}{+}16}-\sqrt{x^2-1}\right)\cdot\frac{\sqrt{x^2\color{red}{+}16}+\sqrt{x^2-1}}{\sqrt{x^2\color{red}{+}16}+\sqrt{x^2-1}}\\
\lim_{x\to\infty}\frac{\left(x^2\color{red}{+}16\right)-\left(x^2-1\right)}{\sqrt{x^2\color{red}{+}16}+\sqrt{x^2-1}}\\
\color{red}{17}\lim_{x\to\infty}\frac{1}{\sqrt{x^2\color{red}{+}16}+\sqrt{x^2-1}}\]
Now the function properly approaches 0 as \(x\to\infty\).
\(\color{blue}{\text{End of Quote}}\)

The result is not affected.

Answer 16

@SithsAndGiggles answer is always going to be zero
any number multiplied with zero is always be zero , there might be some exceptions But i m 99 percent sure it is always going to be zero
the case provided by you exp(x)*exp(-x) is always going to be equal to 1
according to my knowledge
\[\exp(-x) = \frac{ 1 }{ \exp(x) }\]

Answer 17

or one can use basic property \[\exp(x) * \exp(y) = \exp(x+y)\]
after using this property i get 1 directly