# Foundations/Comprehension 1.The following limits are of the form 0/0. Using algebraic

Foundations/Comprehension
1.The following limits are of the form 0/0. Using algebraic methods, simplify the expressions, then compute the limits (5 points each).
a)$$\displaystyle\lim_{{{x}\rightarrow{10}}}{\frac{{{3}{x}^{{{2}}}-{23}{x}-{70}}}{{{11}{x}^{{{2}}}-{1100}}}}$$
b)$$\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{{5}{\left({8}+{h}\right)}^{{{2}}}-{320}}}{{h}}}$$

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Step 1
a)
$$\displaystyle\lim_{{{x}\rightarrow{10}}}{\frac{{{3}{x}^{{{2}}}-{23}{x}-{70}}}{{{11}{x}^{{{2}}}-{1100}}}}$$
As $$\displaystyle{x}\rightarrow{10}$$ then numerator and denominator both tends to zero.
$$\displaystyle{i}\cdot{e}$$ limit is of the form $$\displaystyle{\frac{{{0}}}{{{0}}}}$$ form.
Now $$\displaystyle\lim_{{{x}\rightarrow{10}}}{\frac{{{3}{x}^{{{2}}}-{23}{x}-{70}}}{{{11}{x}^{{{2}}}-{1100}}}}$$
$$\displaystyle=\lim_{{{x}\rightarrow{10}}}{\frac{{{3}{x}^{{{2}}}-{30}{x}+{7}{x}-{70}}}{{{11}{x}^{{{2}}}-{1100}}}}$$
$$\displaystyle=\lim_{{{x}\rightarrow{10}}}{\frac{{{3}{x}{\left({x}-{10}\right)}+{7}{\left({x}-{10}\right)}}}{{{11}{\left({x}^{{{2}}}-{100}\right)}}}}$$ (factoring numerator and denominator)
$$\displaystyle=\lim_{{{x}\rightarrow{10}}}{\frac{{{\left({3}{x}+{7}\right)}{\left({x}-{10}\right)}}}{{{11}{\left({x}-{10}\right)}{\left({x}+{10}\right)}}}}$$ $$\displaystyle{\left(\because{a}^{{{2}}}-{b}^{{{2}}}={\left({a}-{b}\right)}{\left({a}+{b}\right)}\right)}$$
$$\displaystyle=\lim_{{{x}\rightarrow{10}}}{\frac{{{3}{x}+{7}}}{{{11}{\left({x}+{10}\right)}}}}$$ (concelling the common factor)
$$\displaystyle={\frac{{{3}{\left({10}\right)}+{7}}}{{{11}{\left({10}+{10}\right)}}}}$$ (finally compuling the limit)
$$\displaystyle={\frac{{{3}{\left({10}\right)}+{7}}}{{{11}{\left(+{10}\right)}}}}$$
Final fons: $$\displaystyle\lim_{{{x}\rightarrow{10}}}{\frac{{{3}{x}^{{{2}}}-{23}{x}-{70}}}{{{11}{x}^{{{2}}}-{1100}}}}={\frac{{{37}}}{{{220}}}}$$ Ans.
b) $$\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{{5}{\left({8}+{h}\right)}^{{{2}}}-{320}}}{{h}}}$$
As $$\displaystyle{h}\rightarrow{0}$$, both numerator and denominator tends to zero, so limit is of $$\displaystyle{\frac{{{0}}}{{{0}}}}$$ form.
Now $$\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{{5}{\left({8}+{h}\right)}^{{{2}}}-{320}}}{{{h}}}}$$
$$\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{{5}{\left({8}^{{{2}}}+{h}^{{{2}}}+{16}{h}\right)}-{320}}}{{{h}}}}$$
$$\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{\neg{\left\lbrace{320}\right\rbrace}+{5}{h}^{{{2}}}+{16}{h}-\neg{\left\lbrace{320}\right\rbrace}}}{{{h}}}}$$ (simplifying the numerator)
$$\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{{5}{h}^{{{2}}}+{80}{h}}}{{{h}}}}$$
$$\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{{h}{\left({5}{h}+{80}\right)}}}{{{h}}}}$$ (taken common factors out)
$$\displaystyle\lim_{{{h}\rightarrow{0}}}{\left({5}{h}+{80}\right)}$$ (concelling common factor)
$$\displaystyle={5}{\left({0}\right)}+{80}={80}$$ (finally compuling the limit)
$$\displaystyle\therefore\lim_{{{h}\rightarrow{0}}}{\frac{{{5}{\left({8}+{h}\right)}^{{{2}}}-{320}}}{{h}}}={80}$$ Ans
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