Activity 4.
Estimate the value of each of the following limits by constructing appropriate tables of values. Then determine the exact value of the limit by using algebra to simplify the function. Finally, plot each function on an appropriate interval to check your result visually.
(a)
\(\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x-1}\)
Hint.
\((x^2 - 1)\) can be factored.
Answer.
\(2\text{.}\)
Solution.
\(\lim_{x \to 1} \frac{x^2 - 1}{x-1} = \lim_{x \to 1} \frac{(x+1)(x-1)}{x-1} = \lim_{x \to 1} (x+1) = 2\text{.}\)
(b)
\(\displaystyle \lim_{x \to 0} \frac{(2+x)^3 - 8}{x}\)
Hint.
Expand the expression \((2+x)^3\text{,}\) and then combine like terms in the numerator.
Answer.
\(12\text{.}\)
Solution.
\begin{align*}
\lim_{x \to 0} \frac{(2+x)^3 - 8}{x} \amp = \lim_{x \to 0} \frac{8 + 12x + 6x^2 + x^3 - 8}{x} = \lim_{x \to 0} \frac{12x + 6x^2 + x^3}{x}\\
\amp = \lim_{x \to 0} (12 + 6x + x^2) = 12\text{.}
\end{align*}
(c)
\(\displaystyle \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x}\)
Hint.
Try multiplying the given function by this fancy form of 1: \(\frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}\text{.}\)
Answer.
\(\frac{1}{2}\text{.}\)
Solution.
\begin{align*}
\lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \amp = \lim_{x \to 0} \frac{\sqrt{x+1} - 1}{x} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} = \lim_{x \to 0} \frac{x+1-1}{x(\sqrt{x+1}+1)}\\
\amp = \lim_{x \to 0} \frac{1}{\sqrt{x+1}+1} = \frac{1}{2}\text{.}
\end{align*}
