Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 5

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PROBLEM:

Evaluate $\displaystyle \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x}$ .

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SOLUTION:

A straight substitution of $x=0$ leads to the indeterminate form $\frac{0}{0}$ which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

\begin{align*} \lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-9}{2x} & =\:\lim\limits_{x\to 0}\:\frac{\left(x+3\right)^2-\left(3\right)^2}{2x}\\ \\ & =\:\lim\limits_{x\to 0}\:\frac{\left(x+3-3\right)\left(x+3+3\right)}{2x}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x\left(x+6\right)}{2x}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x+6}{2}\\ \\ & =\frac{0+6}{2}\\ \\ & =\frac{6}{2}\\ \\ & =3 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

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