Each limit represents the derivative of some function f at

Redemitz4s 2021-11-24 Answered
Each limit represents the derivative of some function f at some number a. State such an f and a in each case.
\(\displaystyle\lim_{{{t}\rightarrow{1}}}{\frac{{{t}^{{{4}}}-{t}-{2}}}{{{t}-{1}}}}\)

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Expert Answer

Aretha Frazier
Answered 2021-11-25 Author has 7782 answers
Step 1
Consider the limit
\(\displaystyle\lim_{{{t}\rightarrow{1}}}{\frac{{{t}^{{{4}}}-{t}-{2}}}{{{t}-{1}}}}\)
We want to represent it as a derivative of some function f at some number a, \(\displaystyle{\ln{}}\) light of definition 5 we want to write our limit in the form
\(\displaystyle\lim_{{{t}\rightarrow{a}}}{\frac{{{f{{\left({t}\right)}}}-{f{{\left({a}\right)}}}}}{{{t}-{a}}}}\)
Since \(\displaystyle{t}-{1}\) appears in the denominator, and \(\displaystyle{t}\rightarrow{1}\), we are motivated to take \(\displaystyle{a}={1}\)
Looking at the numerator of the limit in question we want \(\displaystyle{f{{\left({t}\right)}}}={t}^{{{4}}}+{t}\). Let us see if this is the right choice. By definition 5 we have
\(\displaystyle{f}'{\left({1}\right)}=\lim_{{{t}\rightarrow{1}}}{\frac{{{f{{\left({t}\right)}}}-{f{{\left({1}\right)}}}}}{{{t}-{1}}}}\)
\(\displaystyle=\lim_{{{t}\rightarrow{1}}}{\frac{{{t}^{{{4}}}+{t}-{\left({1}^{{{4}}}+{1}\right)}}}{{{t}-{1}}}}\)
\(\displaystyle=\lim_{{{t}\rightarrow{1}}}{\frac{{{t}^{{{4}}}+{t}-{2}}}{{{t}-{1}}}}\)
Which is what we wanted to establish.
Ans: \(\displaystyle{f{{\left({t}\right)}}}={t}^{{{4}}}+{t},\ {a}={1}\)
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Hiroko Cabezas
Answered 2021-11-26 Author has 9232 answers
Step 1
Definition
\(\displaystyle{f}'{\left({a}\right)}=\lim_{{{x}\rightarrow{a}}}{\frac{{{f{{\left({x}\right)}}}-{f{{\left({a}\right)}}}}}{{{x}-{a}}}}\)
Step 2
\(\displaystyle\lim_{{{t}\rightarrow{1}}}{\frac{{{t}^{{{4}}}+{t}-{2}}}{{{t}-{1}}}}\)
\(\displaystyle\lim_{{{t}\rightarrow{1}}}{\frac{{{\left({t}^{{{4}}}+{t}\right)}-{\left({1}^{{{4}}}+{1}\right)}}}{{{t}-{1}}}}\)
\(\displaystyle\lim_{{{t}\rightarrow{1}}}{\frac{{{f{{\left({t}\right)}}}-{f{{\left({1}\right)}}}}}{{{t}-{1}}}}={f}'{\left({1}\right)}\)
Where \(\displaystyle{f{{\left({t}\right)}}}={t}^{{{4}}}+{t}\)
Note that \(\displaystyle{a}={1}\)
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