Science:Math Exam Resources/Courses/MATH105/April 2010/Question 06

MATH105 April 2010

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Question 06
Let $f(x)=\lim _{n\to \infty }\left[\left\{{\frac {1}{3}}+{\frac {1}{2+\left(1+{\frac {x-1}{n}}\right)^{3}}}+{\frac {1}{2+\left(1+{\frac {2(x-1)}{n}}\right)^{3}}}+\ldots \right.\right.$ $\ldots +\left.\left.{\frac {1}{2+\left(1+{\frac {(n-1)(x-1)}{n}}\right)^{3}}}\right\}{\frac {x-1}{n}}\right]$ where x ≥ 0. Find the equation of the tangent line to the graph y = f(x) at x = 1.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Try to write the sum above in summation notations (using $\Sigma$). Do you recognize a pattern?

Hint 2
f(x) is the limit of a left Riemann sum and can hence be represented as an integral. Try to find that integral expression.

Hint 3
Recall that the general form of a (left) Riemann sum is $\int _{a}^{b}f(y)\,dy=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}\Delta y\,f(y_{j}),$ where $y_{j}=a+j\Delta y$ and $\Delta y={\frac {b-a}{n}}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can recognize this as the limit of a Riemann sum. Let's first put this into summation notation: $f(x)=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}{\frac {1}{2+(1+j(x-1)/n)^{3}}}\cdot {\frac {x-1}{n}}$ Note that when $j=0$ this gives us the 1/3 term in the sum. We put the factor $(x-1)/n$ first to resemble the general form of a (left) Riemann sum as given in Hint 3. Ultimately, we aim to express $f(x)$ as a definite integral i.e. something of the form $\int _{a}^{b}g(s)ds=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}g(s_{j})\cdot \Delta s$ We choose $s$ here for ease of notation and not overuse the $x$ symbol. For $x$ fixed, we can read off an interval spacing of $\Delta s={\frac {x-1}{n}}$ Let's rewrite the sum with $\Delta s$ in place of $(x-1)/n$: $f(x)=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}{\frac {1}{2+(1+j\Delta s)^{3}}}\cdot \Delta s$. This is suggestive of a left-endpoint Riemann sum, since we start with $j=0$ and end at $j=n-1$. In that case, we can take our subinterval endpoints to be $s_{j}=1+j\Delta s$ (recall that when we do a Riemann sum, we have subintervals with endpoints that look like $s_{j}=a+j\Delta s$ and terms of the form $1+j\Delta s$ are of this form). Let's rewrite this as $f(x)=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}{\frac {1}{2+s_{j}^{3}}}\cdot \Delta s$ From here, we read off $g(s)={\frac {1}{2+s^{3}}}$ Also, since $s_{0}=1+0\Delta s=1$ is the lower bound and $s_{n}=1+n\Delta s=1+n(x-1)/n=x$ is the upper bound, our final result is that $f(x)=\int _{1}^{x}{\frac {1}{2+s^{3}}}ds$ Finally we can compute the equation of the tangent line to $f(x)$ at $x=1$. All we need are $f(1)$ and $f'(1)$. $f(1)=\int _{1}^{1}g(s)ds=0$ since we are integrating over a range of zero. And by the Fundamental Theorem of Calculus, $f'(x)={\frac {1}{2+x^{3}}}$ so $f'(1)={\frac {1}{3}}$ Hence the equation of the tangent line is $y-0={\frac {1}{3}}(x-1)$ or simply $y={\frac {1}{3}}(x-1)$

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