Chapter 11, Problem 31CR

To determine

To find: The limit of the function (if it exists) and confirm the result graphically using a graphical utility.

Answer to Problem 31CR

  $\underset{\mathrm{u}\to 0}{\lim }\frac{\sqrt{4+\mathrm{u}}-2}{\mathrm{u}}=\frac{1}{4}$

Explanation of Solution

Given information: $\underset{\mathrm{u}\to 0}{\lim }\frac{\sqrt{4+\mathrm{u}}-2}{\mathrm{u}}$

Formula used:

• Rationalizing Technique.

To find the limits of some functions is first to rationalize the numerator of the function. Rationalizing the numerator means multiplying the numerator and denominator by the conjugate of the numerator. Then evaluate the limit by direct substitution.

• ${\mathrm{a}}^2-{\mathrm{b}}^2=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}+\mathrm{b}\right)$

Calculation:

  $\underset{\mathrm{u}\to 0}{\lim }\frac{\sqrt{4+\mathrm{u}}-2}{\mathrm{u}}=\underset{\mathrm{u}\to 0}{\lim }\frac{\sqrt{4+\mathrm{u}}-2}{\mathrm{u}}\times \frac{\sqrt{4+\mathrm{u}}+2}{\sqrt{4+\mathrm{u}}+2}$

  $\begin{array}{l}=\underset{\mathrm{u}\to 0}{\lim }\frac{\left(\sqrt{4+\mathrm{u}}\right)-{\left(2\right)}^2}{\mathrm{u}\left(\sqrt{4+\mathrm{u}}+2\right)}\\ =\underset{\mathrm{u}\to 0}{\lim }\frac{4+\mathrm{u}-4}{\mathrm{u}\left(\sqrt{4+\mathrm{u}}+2\right)}\\ \underset{\mathrm{u}\to 0}{\lim }\frac{\mathrm{u}}{\mathrm{u}\left(\sqrt{4+\mathrm{u}}+2\right)}\\ =\underset{\mathrm{u}\to 0}{\lim }\frac{1}{\sqrt{4+\mathrm{u}}+2}\\ =\frac{1}{\sqrt{4+0}+2}\\ =\frac{1}{2+2}\\ =\frac{1}{4}\end{array}$

Graph:

  

Thus, $\underset{\mathrm{u}\to 0}{\lim }\frac{\sqrt{4+\mathrm{u}}-2}{\mathrm{u}}=\frac{1}{4}$