Mathematics: Compute the limit of the following as x approaches infinity. after multiplying both sides by the conjugate, i got the limit 1/∞ after plugging in the limit, but is that the proper answer? 1/∞ would just make the answer 0, right?

Compute the limit of the following as x approaches

karyvillanuevap5w9ym

17.01.2022

Yes, that's the correct limit.

$\sqrt{x^2+1}-x=\dfrac1{\sqrt{x^2+1}+x}=\dfrac1{\sqrt{x^2}\sqrt{1+\frac1{x^2}}+x}$

$\sqrt{x^2}=|x|$ , but since $x\to\infty$ , we are considering , for which |x|=x . Then

$\displaystyle\lim_{x\to\infty}\sqrt{x^2+1}-x=\lim_{x\to\infty}\frac1{x\left(\sqrt{1+\frac1{x^2}}+1\right)}$

and both $\dfrac1x$ and $\dfrac1{x^2}$ vanish as $x\to\infty$ , making the overall limit 0.

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kaelynnmarie1135

03.07.2020

180 - 37 - 37 = ? or 180 - 74 = ?

Step-by-step explanation:

A triangle is equal to 180° so 180 - 37 - 37 or 180 - 74 will give you the missing angle

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