randRange( 4, 7 ) new Polynomial( randRange( 2, 4 ), DEG ) new Polynomial( randRange( 2, 4 ), DEG ) randFromArray([ "", "-" ])

Find \displaystyle\lim_{x \to PM\infty}\dfrac{NUM.text()}{DEN.text()}.

fractionReduce( NUM.getCoefAndDegreeForTerm(0).coef, DEN.getCoefAndDegreeForTerm(0).coef )

• fractionReduce( -NUM.getCoefAndDegreeForTerm(0).coef, DEN.getCoefAndDegreeForTerm(0).coef )
• +\infty
• -\infty
• 0
• undefined

Look at the leading terms expr(NUM.expr()[1]) and expr(DEN.expr()[1]).

Because they have the same degree DEG, the limit is equal to the quotient of their coefficients.

\displaystyle\lim_{x \to PM\infty}\dfrac{NUM.text()}{DEN.text()} = fractionSimplification( NUM.coefs[DEG], DEN.coefs[DEG] )

randRange( 4, 7 ) new Polynomial( randRange( 2, 4 ), randRange( 4, 7 ) ) new Polynomial( randRange( 2, 4 ), randRange( 4, 7 ) ) randFromArray([ "", "-" ])

Find \displaystyle\lim_{x \to PM\infty}\dfrac{NUM.text()}{DEN.text()}.

0

• fractionReduce( NUM.getCoefAndDegreeForTerm(0).coef, DEN.getCoefAndDegreeForTerm(0).coef )
• fractionReduce( -NUM.getCoefAndDegreeForTerm(0).coef, DEN.getCoefAndDegreeForTerm(0).coef )
• +\infty
• -\infty
• undefined

Look at the leading terms expr(NUM.expr()[1]) and expr(DEN.expr()[1]).

Because the numerator's degree NUM.getCoefAndDegreeForTerm(0).degree is less than the denominator's degree DEN.getCoefAndDegreeForTerm(0).degree, the bottom term dominates as x approaches PM\infty.

Since the denominator grows faster than the numerator, the limit goes to 0.

randRange( 4, 7 ) new Polynomial( randRange( 2, 4 ), randRange( 4, 7 ) ) new Polynomial( randRange( 2, 4 ), randRange( 4, 7 ) ) NUM.getCoefAndDegreeForTerm(0).coef * DEN.getCoefAndDegreeForTerm(0).coef > 0 ? "+" : "-" RIGHT_SIGN === "+" ? "-" : "+"

Find \displaystyle\lim_{x \to \infty}\dfrac{NUM.text()}{DEN.text()}.

RIGHT_SIGN\infty

• fractionReduce( NUM.getCoefAndDegreeForTerm(0).coef, DEN.getCoefAndDegreeForTerm(0).coef )
• fractionReduce( -NUM.getCoefAndDegreeForTerm(0).coef, DEN.getCoefAndDegreeForTerm(0).coef )
• WRONG_SIGN\infty
• 0
• undefined

Look at the leading terms expr(NUM.expr()[1]) and expr(DEN.expr()[1]).

As x \to \infty, the numerator approaches NUM.getCoefAndDegreeForTerm(0).coef < 0 ? "-" : ""\infty because the coefficient NUM.getCoefAndDegreeForTerm(0).coef is NUM.getCoefAndDegreeForTerm(0).coef < 0 ? "negative" : "positive".

As x \to \infty, the denominator NUM.getCoefAndDegreeForTerm(0).coef * DEN.getCoefAndDegreeForTerm(0).coef > 0 ? "also " : ""approaches DEN.getCoefAndDegreeForTerm(0).coef < 0 ? "-" : ""\infty because the coefficient DEN.getCoefAndDegreeForTerm(0).coef is DEN.getCoefAndDegreeForTerm(0).coef < 0 ? "negative" : "positive".

Because the numerator's degree NUM.getCoefAndDegreeForTerm(0).degree is greater than the denominator's degree DEN.getCoefAndDegreeForTerm(0).degree, the limit diverges.

The numerator and denominator have the same sign as x gets large, so the limit is +\infty.

The numerator and denominator have differing signs as x gets large, so the limit is -\infty.

randRangeNonZero( -7, 7 ) randRangeNonZero( -7, 7 ) A > 0 ? "-" : "+" SIGN_LIM_LEFT === "+" ? "-" : "+"

Find \displaystyle\lim_{x \to K}\dfrac{A}{Bx + -K}.

undefined

• fractionReduce( A, K )
• fractionReduce( A, -K )
• +\infty
• -\infty
• 0

Consider the behavior of the function as x \to K from each direction.

As x approaches K from the left, Bx + -K starts negative and increases as it approaches 0, so \dfrac{A}{Bx + -K} approaches SIGN_LIM_LEFT\infty.

As x approaches K from the right, Bx + -K starts positive and decreases as it approaches 0, so \dfrac{A}{Bx + -K} approaches SIGN_LIM_RIGHT\infty.

Since the left- and right-hand limits are not equal, the limit is not defined.

randRangeNonZero( -7, 7 ) randRangeNonZero( -7, 7 ) A > 0 ? "+" : "-" RIGHT_SIGN === "+" ? "-" : "+"

Find \displaystyle\lim_{x \to K}\dfrac{A}{(Bx + -K\smash{)}^2}.

RIGHT_SIGN\infty

• fractionReduce( A, K * K )
• fractionReduce( A, -K * K )
• WRONG_SIGN\infty
• 0
• undefined

Consider the behavior of the function as x \to K from each direction.

In either direction, (x + -K)^2 approaches 0, so \dfrac{A}{(Bx + -K\smash{)}^2} diverges.

Because (x + -K)^2 is always positive and A is A > 0 ? "positive" : "negative", \dfrac{A}{(Bx + -K\smash{)}^2} approaches RIGHT_SIGN\infty.