High School

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Given that lim p(x) = 4 and lim q(x) = -3. Find:

a) lim (3p(x) - x) as x→1

b) lim (5q(x) + 4) as x→1

c) lim (-p(x)q(x)) as x→3

d) lim (\frac{\sqrt{4p(x)}}{-2}) as x→2

Answer :

Let's solve the given limits step-by-step by applying the properties of limits. We are provided with [tex]\lim p(x) = 4[/tex] and [tex]\lim q(x) = -3[/tex]. These are the limits of the functions [tex]p(x)[/tex] and [tex]q(x)[/tex] at any point [tex]x[/tex], and we'll use this information to find the following limits:

(a) [tex]\lim_{x \to 1} (3p(x) - x)[/tex]

We can separate the limit and calculate:
[tex]\lim_{x \to 1} (3p(x) - x) = \lim_{x \to 1} (3p(x)) - \lim_{x \to 1} (x)[/tex]

Using properties of limits, we have:
[tex]\lim_{x \to 1} (3p(x)) = 3 \cdot \lim_{x \to 1} (p(x)) = 3 \cdot 4 = 12[/tex]
[tex]\lim_{x \to 1} (x) = 1[/tex]
Thus,
[tex]\lim_{x \to 1} (3p(x) - x) = 12 - 1 = 11[/tex]

(b) [tex]\lim_{x \to 1} (5q(x) + 4)[/tex]

Similarly, we separate the limit:
[tex]\lim_{x \to 1} (5q(x) + 4) = \lim_{x \to 1} (5q(x)) + \lim_{x \to 1} (4)[/tex]

Using properties of limits, we have:
[tex]\lim_{x \to 1} (5q(x)) = 5 \cdot \lim_{x \to 1} (q(x)) = 5 \cdot (-3) = -15[/tex]
[tex]\lim_{x \to 1} (4) = 4[/tex]
Thus,
[tex]\lim_{x \to 1} (5q(x) + 4) = -15 + 4 = -11[/tex]

(c) [tex]\lim_{x \to 3} (-p(x)q(x))[/tex]

First, find [tex]\lim_{x \to 3} (p(x)q(x))[/tex]:
[tex]\lim_{x \to 3} (p(x)q(x)) = \lim_{x \to 3} (p(x)) \cdot \lim_{x \to 3} (q(x)) = 4 \cdot (-3) = -12[/tex]
Thus,
[tex]\lim_{x \to 3} (-p(x)q(x)) = -(-12) = 12[/tex]

(d) [tex]\lim_{x \to 2} \left( \frac{\sqrt{4p(x)}}{-2} \right)[/tex]

First, evaluate the limit inside the square root:
[tex]\lim_{x \to 2} \sqrt{4p(x)} = \sqrt{4 \times \lim_{x \to 2} p(x)} = \sqrt{4 \times 4} = \sqrt{16} = 4[/tex]
Then, apply this to the original expression:
[tex]\lim_{x \to 2} \left( \frac{\sqrt{4p(x)}}{-2} \right) = \frac{4}{-2} = -2[/tex]

In conclusion, the limits are:


  1. [tex]\lim_{x \to 1} (3p(x) - x) = 11[/tex]

  2. [tex]\lim_{x \to 1} (5q(x) + 4) = -11[/tex]

  3. [tex]\lim_{x \to 3} (-p(x)q(x)) = 12[/tex]

  4. [tex]\lim_{x \to 2} \left( \frac{\sqrt{4p(x)}}{-2} \right) = -2[/tex]

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