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Mean
Mode
Greatest Common Factor
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Order of Operations
Fractions
Mixed Fractions
Prime Factorization
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Algebra
Combine Like Terms
Solve for a Variable
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Expand
Evaluate Fractions
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Simplify
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Evaluate
\frac{3}{2}=1.5
Quiz
Limits
5 problems similar to:
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
Similar Problems from Web Search
What is the value of \displaystyle\lim_{{{x}\rightarrow-{3}}}{\frac{{{x}^{{{2}}}-{9}}}{{{x}^{{{2}}}-{2}{x}-{15}}}} ?
https://socratic.org/questions/what-is-the-value-of-lim-x-rightarrow-3-frac-x-2-9-x-2-2x-15
\displaystyle\frac{{3}}{{4}} Explanation: \displaystyle\lim_{{{x}\to-{3}}}\frac{{{x}^{{2}}-{9}}}{{{x}^{{2}}-{2}{x}-{15}}} By factoring out the numerator and the denominator, \displaystyle=\lim_{{{x}\to-{3}}}\frac{{\cancel{{{\left({x}+{3}\right)}}}{\left({x}-{3}\right)}}}{{\cancel{{{\left({x}+{3}\right)}}}{\left({x}-{5}\right)}}}=\frac{{-{3}-{3}}}{{-{3}-{5}}}=\frac{{-{6}}}{{-{8}}}=\frac{{3}}{{4}} ...
Evaluating \lim\limits_{x\to-3}\frac{x^2-9}{2x^2+7x+3}
https://math.stackexchange.com/q/490053
\require{cancel} Note that both the numerator and denominator evaluate to 0 at x = -3. That means we have an indeterminate limit, which simply means more work needs to be done. Factor ...
Help with epsilon-delta proof of \lim_{x \to -3} \frac{x^2 + x - 6}{x^2 - 9}
https://math.stackexchange.com/questions/197844/help-with-epsilon-delta-proof-of-lim-x-to-3-fracx2-x-6x2-9
HINT: \frac{x^2+x-6}{x^2-9}-\frac56=\frac{6x^2+6x-36-(5x^2-45)}{6(x^2-9)}=\frac{x^2+6x+9}{6(x^2-9)}=\frac{(x+3)^2}{6(x-3)(x+3)}
What is \displaystyle \lim_{x \to 1} \dfrac{x^2-1}{x^2+3x-4} ?
https://www.quora.com/What-is-displaystyle-lim_-x-to-1-dfrac-x-2-1-x-2+3x-4
You can see if you solve for x=1 that you get a fraction in the form of \dfrac{0}{0} which is indeterminate. When you get into a situation like this you can apply L'Hôpital's rule - Wikipedia. ...
Proving that \lim_{x\to 2}\frac{x^2-5x}{x^2+2}=-1 using the \epsilon-\delta definition of a limit
https://math.stackexchange.com/questions/1872290/proving-that-lim-x-to-2-fracx2-5xx22-1-using-the-epsilon-delta
Let \epsilon>0 be given. We have to find a number \delta>0 such that \left|\frac{x^2-5x}{x^2+2}+1\right|<\epsilon whenever |x-2|<\delta. But, as Andre notes, \left|\frac{x^2-5x}{x^2+2}+1\right|=\left|\frac{2x^2-5x+2}{x^2+2}\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|. ...
Limit simplification
https://math.stackexchange.com/q/2704118
Note that \lim_{x\to -3} \frac{(x^2-9)^2}{x^2+\color{red}6x+9} We can solve in this way \lim_{x\to -3} \frac{(x^2-9)^2}{(x+3)^2}=\lim_{x\to -3} \frac{(x-3)^2(x+3)^2}{(x+3)^2}=\lim_{x\to -3} (x-3)^2=36
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\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}
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