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Possible Duplicate:
Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$

$$\lim_{x \to \infty} \sqrt{4x^2 + 4} - (2x + 2)$$

So, I have an intermediate form of $\infty - \infty$ and I tried multiplying by the conjugate; however, I seem to be left with another intermediate form of $\frac{\infty}{\infty}$ and wasn't sure what else to to do. Is there anything else I can do other than L'Hopital's rule?

Community
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stariz77
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  • Once it is in the form $\frac{\infty}{\infty}$, divide top and bottom by the highest "rate of growth" of any term that you see. In this case, this will allow you to avoid L'Hopital's rule (assuming you know a few basic limits). – Barry Smith Mar 28 '12 at 23:21

3 Answers3

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$$\begin{align} \sqrt{4x^2 + 4} - (2x+2) & = \frac{(\sqrt{4x^2 + 4} - (2x+2))(\sqrt{4x^2 + 4} + (2x+2))}{\sqrt{4x^2 + 4} + (2x+2)}\\ & = \frac{4x^2 +4 - 4(x+1)^2}{\sqrt{4x^2 + 4} + (2x+2)} = - \frac{8x}{\sqrt{4x^2 + 4} + (2x+2)}\\ & = - \frac{8}{\sqrt{4 + 4/x^2} + 2 + 2/x} \end{align} $$ Now take the limit as $x \rightarrow \infty$ to get the limit as $-2$.

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Hint: $$ \sqrt{4x^2+4}-(2x+2)=\frac{-8x}{\sqrt{4x^2+4}+(2x+2)}=\frac{-8}{\operatorname{sign}(x)\sqrt{4+\frac{4}{x^2}}+\left(2+\frac{2}{x}\right)} $$

Norbert
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Multiplying and dividing by $\sqrt{4x^2 + 4} + (2x + 2)$ you obtain: $$\begin{split} \sqrt{4x^2 + 4} - (2x + 2) &= \frac{(\sqrt{4x^2 + 4} - (2x + 2))(\sqrt{4x^2 + 4} + (2x + 2))}{\sqrt{4x^2 + 4} + (2x + 2)} \\ &= \frac{(4x^2 + 4) - (2x+2)^2}{\sqrt{4x^2 + 4} + (2x + 2)} \\ &= \frac{-8x}{\sqrt{4x^2 + 4} + (2x + 2)} \\ &= \frac{-8x}{2x (\sqrt{1+\frac{1}{x^2}} + 1 + \frac{1}{x})} \\ &= - \frac{4}{\sqrt{1+\frac{1}{x^2}} + 1 + \frac{1}{x}} \end{split}$$ hence the limit equals $-2$.

Pacciu
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