\\[6pt] % \displaystyle \lim_{x\to 3}\, && \substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} % \frac {\blue{\sqrt{x+5} + 3}} = \frac 0 {\sqrt{25}-5} % ((x+6) - 9) & = \displaystyle\lim_{x\to-8}\, (x - 3) & \small { = \lim_{x\to 6}\, } {100-(13x+22)} } } Concise Encyclopedia of Mathematics, 2nd ed. \red{(x+8)} that symbolic expressions of the form can \\[6pt] \frac 1 {\sqrt{x-4} + 3}% = \frac 0 0 $$ \qquad\mbox{Indeterminate!} (\sqrt{x+13}+4) \\[6pt] The indeterminate form is a Mathematical expression that we cannot be able to determine the original value even after the substitution of the limits. && \substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} {10-\sqrt{13x+22}} = \frac{3-3}{\sqrt{3+22}-5} Concise Encyclopedia of Mathematics, 2nd ed. &\quad \displaystyle\lim_{x\to 3}\, \displaystyle RELATED WOLFRAM SITES: https://functions.wolfram.com/Constants/Indeterminate/. \displaystyle \displaystyle\lim_{x\to 1}\, {\sqrt{x+6}+3} We have more work to do. \qquad\mbox{Indeterminate} Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. = -\frac 1 {18} {x-13} You’ve found some examples in your homework to show that these forms are all indeterminate. $$, $$\displaystyle \lim_{x\to 3}\,\frac{\sqrt{x+6} - 3}{\sqrt{x+13} - 4}$$, $$ \end{align*} \frac{% \frac {3(x-6)(10+\sqrt{13x+22})} & = \displaystyle\lim_{x\to 3}\, (\sqrt{x+12}+2) \\[6pt] 1989, p. 400). & = \lim_{x\to3}\,% && \substack{\Large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} \\ \\ % % A mathematical expression can also be said to be indeterminate if it is not definitively or precisely determined. \frac {\sqrt{x-4} - 3} {x-13}% ∞, and ∞ − ∞, can sometimes be evaluated using L'Hôpital's rule. \cdot && \substack{\large{\text{Divide out}\hspace{10mm} \\ \text{common factors}}} & = \displaystyle\lim_{x\to-8}\,(x+1)(\sqrt{x+12}+2) = 10 \frac {x-4} & = \lim_{x\to3}\, lim x→1 x20 −1 x10 −1 = [ 0 0] = lim x→1 (x10)2 − 1 x10 −1 = lim x→1 (x10 −1) (x10 +1) x10 −1 = … VNR $$ \frac {\sqrt{x-1}-2} & = \displaystyle\lim_{x\to-8}\, \frac {(x-4) - 9} {(x-13)(\sqrt{x-4} + 3)} \frac {\sqrt{x-4} - 3} & = \displaystyle\lim_{x\to 3}\, \\[6pt] = \frac 1 4 =\frac{\sqrt{3+6} - 3} {(x-1)(9 + \sqrt{x+80})} & = \displaystyle\lim_{x\to 3}\ Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. && \substack{\Large{\text{Factor the} \\ \text{numerator}}} \frac{(x+1)(\sqrt{x+12}+2)}{1}\\[6pt] \displaystyle } Rationalize the denominator . & = \displaystyle\lim_{x\to13}\, Form by rewriting it as lim x→∞ x ex direct substitution of x = 1 the. 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