MHB What is the simplified form of the limit as x approaches infinity?

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The limit as x approaches infinity for the expression \(\lim_{x\to +\infty}\sqrt{x^2+3}-x\) can be simplified by multiplying by the conjugate \(\frac{\sqrt{x^2+3} + x}{\sqrt{x^2+3} + x}\). This approach helps eliminate the indeterminate form. Alternatively, using the Binomial Theorem, the limit can be expressed as \(\lim_{x\to +\infty}\left(x + \frac{3}{2x} - x\right)\), which simplifies further. Ultimately, the limit evaluates to \(\frac{3}{2}\). Understanding these methods is crucial for solving similar limits effectively.
karush
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\lim_{x\to +\infty}\sqrt{x^2+3}-x

sorry first of all how do you turn this string into latex
I don't see the icon tool on the editor

thnx
 
Last edited by a moderator:
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karush said:
\lim_{x\to +\infty}\sqrt{x^2+3}-x}

sorry first of all how do you turn this string into latex
I don't see the icon tool on the editor

thnx
$$\lim_{x \to +\infty} \sqrt{x^2+3}-x$$

Click on Reply With Quote to see the LaTex. Or better yet, there is a "How To Use LaTex on This Site" thread in the LaTeX Help section of the site.

Hint: Try multiplying by $$ \frac{\sqrt{x^2+3} + x}{\sqrt{x^2+3} + x} $$
 
Last edited:
karush & edit said:
\lim_{x\to +\infty}{(\sqrt{x^2 + 3} - x)}

Redo post:Alternative
Using the Binomial Theorem for (x^2 + 3)^{\frac{1}{2}},

write the limit as \displaystyle\lim_{x\to +\infty}{\bigg(x + \dfrac{3}{2x} + (all \ \ other \ \ terms \ \ with \ \ degree \ \ less \ \ than \ \ -1) \ - \ x \bigg)}
 
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