7. Computing Limits

d. Limits at Infinity

3. When Limit Laws Don't Apply

Limits without Laws - Limit Tricks

c. Multiply by the Conjugate

The conjugate of a difference \(a-b\) is the sum \(a+b\). This is useful because their product is the difference of squares, \((a-b)(a+b)=a^2-b^2\). Consequently, if \(a\) and/or \(b\) is a square root, then \(a^2-b^2\) eliminates the square root.

This trick says if numerator or denominator contains a difference where one or both terms is a square root, then multiply the numerator and denominator by the conjugate. Multiply out the numerator or denominator which had the original difference. It is most appropriate for the indeterminate forms \(\dfrac{0}{0}\), \(\dfrac{\infty}{\infty}\), \(0\cdot\infty\) or \(\infty-\infty\).

Compute \(\lim\limits_{x\to\infty} \left(\sqrt{x+\sqrt{x}\;}-\sqrt{x}\right)\).

This limit has the form \(\infty-\infty\). We multiply and divide by the conjugate, \(\sqrt{x+\sqrt{x}}+\sqrt{x}\): \[\begin{aligned} \lim_{x\to\infty} &\left(\sqrt{x+\sqrt{x}\;}-\sqrt{x}\right) \\ &=\lim_{x\to\infty} \dfrac{\left(\sqrt{x+\sqrt{x}\;}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}\;}+\sqrt{x}\right)} {\left(\sqrt{x+\sqrt{x}\;}+\sqrt{x}\right)} \\ &=\lim_{x\to\infty} \dfrac{x+\sqrt{x}-x} {\sqrt{x+\sqrt{x}\;}+\sqrt{x}} \\ &=\lim_{x\to\infty} \dfrac{\sqrt{x}} {\sqrt{x+\sqrt{x}\;}+\sqrt{x}} \\ &=\lim_{x\to\infty} \dfrac{1} {\sqrt{1+\dfrac{1}{\sqrt{x}}}+1}=\dfrac{1}{2} \end{aligned}\] In the next to last step, we divided the numerator and denominator by \(\sqrt{x}\).