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Evaluate each of the following limits: a) lim(z->2) (z3 + 2z - 6) b) lim(z->infinity) (4z2 + 3z - 2)/(z2 + 3z + 1)

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2 parts
a

6\text{6}

b

4\text{4}

Step-by-step solution

7 steps

let's keep going with the next part of your exercise!

We're going to "evaluate each of the following," which means finding the value of these expressions as zz gets closer and closer to a certain number, or very, very big.

2. Evaluate each of the following.

Part a· 3 steps

lim⁡z→2(z3+2z−6)\lim_{z \to 2} (z^3 + 2z - 6)

Imagine you have a smooth road, and you want to know the height of the road when you're exactly at mile marker 2. Since this road (function) is smooth, you can just plug in the mile marker number to find the height!

  1. Substitute z=2z=2 into the expression.

    lim⁡z→2(z3+2z−6)=(2)3+2(2)−6\lim_{z \to 2} (z^3 + 2z - 6) = (2)^3 + 2(2) - 6

  2. Calculate the values.

    =8+4−6= 8 + 4 - 6

  3. Simplify the expression.

    =12−6= 12 - 6 =6= 6

    The limit is 6\boxed{\text{6}}.

Part b

4\text{4}

lim⁡z→∞4z2+3z−2z2+3z+1\lim_{z \to \infty} \frac{4z^2 + 3z - 2}{z^2 + 3z + 1}

Now, imagine you're flying very high in an airplane, looking down at two very tall buildings. You want to know which building is taller, but they're both so huge that the little details (like a flag on top or a small balcony) don't really matter. What matters most is the main height of each building.

In this math problem, when zz goes to "infinity" (∞\infty), it means zz is becoming incredibly, incredibly large. When zz is huge, the terms with the highest power of zz (like z2z^2) become much, much more important than the terms with smaller powers of zz (like zz or just a number).

  1. Identify the highest power of zz in the denominator.

    The highest power of zz in the denominator (z2+3z+1z^2 + 3z + 1) is z2z^2.

  2. Divide every term in the numerator and the denominator by z2z^2.

    lim⁡z→∞4z2z2+3zz2−2z2z2z2+3zz2+1z2\lim_{z \to \infty} \frac{\frac{4z^2}{z^2} + \frac{3z}{z^2} - \frac{2}{z^2}}{\frac{z^2}{z^2} + \frac{3z}{z^2} + \frac{1}{z^2}}

  3. Simplify the terms.

    Remember that as z→∞z \to \infty, any number divided by zz (or z2z^2, etc.) goes to 0.

    lim⁡z→∞4+3z−2z21+3z+1z2\lim_{z \to \infty} \frac{4 + \frac{3}{z} - \frac{2}{z^2}}{1 + \frac{3}{z} + \frac{1}{z^2}}

  4. Evaluate the limit.

    =4+0−01+0+0= \frac{4 + 0 - 0}{1 + 0 + 0} =41= \frac{4}{1} =4= 4

    The limit is 4\boxed{\text{4}}.

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