Limits by conjugates

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When we want to evaluate a limit of a function, it is sometimes useful to know the rationalizing the numerator/denominator of the fraction in the function itself.

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

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The conjugate of \( \sqrt{x^2+2x} - x \) is \( \sqrt{x^2+2x} + x \), so we multiply the limit by the function \( \dfrac{\sqrt{x^2+2x} + x}{\sqrt{x^2+2x} + x} \) to get

\[ \begin{eqnarray} \lim_{x\to\infty}\left( \sqrt{x^2+2x} - x \right) & = &\lim_{x\to\infty} \left( \sqrt{x^2+2x} - x \right) \cdot\dfrac{\sqrt{x^2+2x} + x}{\sqrt{x^2+2x} + x} \\ &=& \lim_{x\to\infty}\dfrac{ \left( \sqrt{x^2+2x} - x \right) \left( \sqrt{x^2+2x} + x \right)} {\sqrt{x^2+2x} + x} \\ &=& \lim_{x\to\infty}\dfrac{ (x^2+2x) - x^2} {\sqrt{x^2+2x} + x} \\ &=& \lim_{x\to\infty}\dfrac{ 2x} {\sqrt{x^2+2x} + x} \\ &=& \lim_{x\to\infty}\dfrac{ 2} {\sqrt{1+2/x} + 1} \\ &=& \dfrac2{\sqrt{1+0} + 1} = 1 \end{eqnarray}\]

• Limits by Taylor series

• L'Hôpital's rule

• Limits by factoring

• Indeterminate forms