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

defazajx 2021-09-22 Answered
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)\(\displaystyle\lim_{{{h}\rightarrow{0}}}{\frac{{{5}{\left({8}+{h}\right)}^{{{2}}}-{320}}}{{h}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Plainmath recommends

  • Ask your own question for free.
  • Get a detailed answer even on the hardest topics.
  • Ask an expert for a step-by-step guidance to learn to do it yourself.
Ask Question

Expert Answer

Arnold Odonnell
Answered 2021-09-23 Author has 8924 answers
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
Have a similar question?
Ask An Expert
43
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

...