Limit practice problems

1. 1
   Use continuity

   Polynomial limits equal their function values.

   $$\lim_{x\to2}(x^3-4x+1)=2^3-4(2)+1$$
2. 2
   Simplify

   Evaluate the arithmetic carefully.

   $$8-8+1=1$$

Independent check: Values at 1.99 and 2.01 are close to 1 from opposite sides.

1. 1
   Factor the numerator

   Use a difference of squares.

   $$x^2-16=(x-4)(x+4)$$
2. 2
   Simplify near 4

   The cancelled expression agrees away from the hole.

   $$\frac{(x-4)(x+4)}{x-4}=x+4$$
3. 3
   Substitute

   The simplified function is continuous at 4.

   $$4+4=8$$

Independent check: The original quotient equals x + 4 for every x near 4 except 4 itself.

1. 1
   Use the conjugate

   The numerator becomes x minus 4.

   $$\frac{\sqrt{x+5}-3}{x-4}\frac{\sqrt{x+5}+3}{\sqrt{x+5}+3}=\frac{1}{\sqrt{x+5}+3}$$
2. 2
   Take the limit

   Now direct substitution is valid.

   $$\frac{1}{3+3}=\frac16$$

Independent check: The quotient is a secant slope for √(x + 5), whose derivative at 4 is 1/(2·3) = 1/6.

1. 1
   Create the standard form

   Isolate sine u divided by u.

   $$\frac{\sin(5x)}{x}=5\frac{\sin(5x)}{5x}$$
2. 2
   Apply the standard limit

   The bracketed quotient approaches 1 in radians.

   $$5\cdot1=5$$

Independent check: High-precision values at x = ±0.0001 agree with 5 to several decimal places.

1. 1
   Rescale

   Create the standard form with u = 2x.

   $$\frac{e^{2x}-1}{x}=2\frac{e^{2x}-1}{2x}$$
2. 2
   Use the standard limit

   As 2x approaches zero, the quotient approaches 1.

   $$2\cdot1=2$$

Independent check: This limit is the derivative of e^(2x) at x = 0, which the chain rule gives as 2.

1. 1
   Select the left-side rule

   For x less than zero, absolute value x equals negative x.

   $$|x|=-x$$
2. 2
   Simplify

   The quotient is constant on the left of zero.

   $$\frac{x}{|x|}=\frac{x}{-x}=-1$$

Independent check: Every negative test input, including -0.1 and -0.001, gives exactly -1.

1. 1
   Approach from the left

   Use x plus 1 for inputs below 2.

   $$\lim_{x\to2^-}f(x)=2+1=3$$
2. 2
   Approach from the right

   Use 5 minus x for inputs at or above 2.

   $$\lim_{x\to2^+}f(x)=5-2=3$$
3. 3
   Compare

   Matching one-sided limits establish the two-sided limit.

   $$\lim_{x\to2}f(x)=3$$

Independent check: The values at 1.999 and 2.001 are 2.999 on both sides.

1. 1
   Normalize by the highest power

   Divide numerator and denominator by x squared.

   $$\frac{3-1/x}{2+5/x^2}$$
2. 2
   Take the limit

   Both reciprocal terms approach zero.

   $$\frac{3-0}{2+0}=\frac32$$

Independent check: For x = 10,000 the quotient is within 0.0001 of 1.5.

1. 1
   Approach from the left

   Negative denominators produce decreasing unbounded values.

   $$\lim_{x\to0^-}\frac1x=-\infty$$
2. 2
   Approach from the right

   Positive denominators produce increasing unbounded values.

   $$\lim_{x\to0^+}\frac1x=+\infty$$
3. 3
   Compare

   The one-sided behaviors do not agree.

   $$\lim_{x\to0}\frac1x\text{ does not exist}$$

Independent check: Reciprocal values at ±0.001 have equal magnitudes and opposite signs.

1. 1
   Bound the oscillation

   Sine always lies between negative one and one.

   $$-1\le\sin(1/x)\le1$$
2. 2
   Multiply by x squared

   The nonnegative factor preserves the order.

   $$-x^2\le x^2\sin(1/x)\le x^2$$
3. 3
   Squeeze

   Both outside functions approach zero.

   $$\lim_{x\to0}x^2\sin(1/x)=0$$

Independent check: The absolute value is at most x², which tends to zero regardless of the oscillation.