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 -\infty because the coefficient NUM.getCoefAndDegreeForTerm(0).coef is negative.

As x \to \infty, the numerator approaches \infty because the coefficient NUM.getCoefAndDegreeForTerm(0).coef is positive.

As x \to \infty, the denominator also approaches -\infty because the coefficient DEN.getCoefAndDegreeForTerm(0).coef is negative.

As x \to \infty, the denominator also approaches \infty because the coefficient DEN.getCoefAndDegreeForTerm(0).coef is 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}{x + -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, x + -K starts negative and increases as it approaches 0, so \dfrac{A}{x + -K} approaches SIGN_LIM_LEFT\infty.

As x approaches K from the right, x + -K starts positive and decreases as it approaches 0, so \dfrac{A}{x + -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}{(x + -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}{(x + -K\smash{)}^2} diverges.

Because (x + -K)^2 is always positive and A is positive, \dfrac{A}{(x + -K\smash{)}^2} approaches RIGHT_SIGN\infty.

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