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  • Functions
    • Polynomials
    • Rational Functions
    • Logarithms
    • Exponentials
    • Trigometric

  • Limits
    • Direct Substitution Property
    • Undetermined Forms
    • Trigometric Limits
    • Factorization
    • Side Limits
    • L'Hopital Rule

  • Continuity
    • Connection with limits
    • Algebraic Properties
    • Intermediate Value Theorem

  • Derivatives
    • Tangent Line
    • Product Rule
    • Quotient Rule
    • Chain Rule
    • Related Rates
    • Mean Value Theorem
    • Maximum and Minimum values

  • Integrals
    • Antiderivatives
    • Fundamental Theorem of Calculus
    • Substitutions and change of variables
    • Partial Fractions
    • Improper Integrals

  • Series
    • Positive Series
    • Convergence criteria
    • Alternating Series





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SAMPLE CALCULUS PROBLEMS

Problem 1. Find

$\displaystyle \lim_{n\rightarrow \infty} \frac{2n^2+1}{3n^2-5n+2}$

Solution.

The problem presents the typical case of a quotient of two quantities that go to infinity. In fact, $ 2n^2+1$ and $ 3n^2-5n+2$ both approach to infinity as $ n\rightarrow\infty$ . Which means that the quotient

$\displaystyle \frac{2n^2+1}{3n^2-5n+2}$

is an undetermined form. Somehow we have to cancel the "bad" part, if possible. Notice that since both $ 2n^2+1$ and $ 3n^2-5n+2$ are big, there's a chance for cancelation, so the quotient could be finite.

The trick in this case is to divide by the highest power on the expression (from both numerator and denominator). In this case we will divide both numerator and denominator by $ n^2$ . We get

$\displaystyle \frac{2n^2+1}{3n^2-5n+2}= \frac{2+\frac{1}{n^2}}{ 3-\frac{5}{n}+\frac{2}{n^2} }\rightarrow \frac{2}{3}$

as $ n\rightarrow\infty$ because $ \frac{1}{n^2}\rightarrow 0$ , $ \frac{5}{n}\rightarrow 0$ and $ \frac{2}{n^2} \rightarrow 0$ as $ n\rightarrow\infty$ .

This trick is applied in general for this type of expressions. $ \Box$



Problem 2. Find $ \frac{dy}{dx}$ and $ \frac{d^2 y}{dx^2}$ if

$\displaystyle x^{2/3}+y^{2/3}=a^{2/3}, \qquad a>0 $


Solution. Let's differentiate both sides of the equality. Let's recall that we are assuming that $ y=y(x)$ . That means that the equation gives us a way to solve $ y$ as a function of $ x$ , but we still don't know how to express that function. That assumption requires further justification (meaning, it's not always granted), but we'll take it for granted. So, since $ y=y(x)$ we get using the chain rule:

$\displaystyle \frac{2}{3} x^{-1/2}+\frac{2}{3} y^{-1/3} y'=0 $

where $ y'\equiv \frac{dy}{dx}$ . The 0 on the right hand side comes from differentiating the constant $ a^{2/3}$ . We multiply the equation by $ 3/2$ and we get

$\displaystyle x^{-1/2}+ y^{-1/3} y'=0 $

which means that

$\displaystyle \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} $

Let's get $ y''$ . From the last equation we obtain easily that

$\displaystyle x\left(\frac{dy}{dx}\right)^3=-y $

and now we differentiate this again with respect to $ x$ to get:

$\displaystyle \left(\frac{dy}{dx}\right)^3+3x\left(\frac{dy}{dx}\right)^2 y''=-y'
							$

But $ \left(\frac{dy}{dx}\right)^3=-y/x$ and $ 3x\left(\frac{dy}{dx}\right)^2=3x(y/x)^{2/3}$ . We know put all the elements together:

$\displaystyle y''=\frac{y/x+(y/x)^{1/3}}{3x(y/x)^{2/3}} $

and now we simplify the right hand side of the last equation

$\displaystyle y''=\frac{(y/x)^{2/3}+1}{3x(y/x)^{1/3}}=\frac{y^{2/3}+x^{2/3}}{3x^{4/3}y^{1/3}} $

But by definition: $ x^{2/3}+y^{2/3}=a^{2/3}$ , so we replace this into the previous equation:

$\displaystyle y''=\frac{a^{2/3}}{3x^{4/3}y^{1/3}} $

which gives the final answer.$ \Box$