# Each limit represents the derivative of some function f at

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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Aretha Frazier
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
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 $$\displaystyle{a}={1}$$