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How to solve limxo2(x2+3x−5)lim_{x o 2} (x^2 + 3x - 5)limxo2​(x2+3x−5)?

  1. Direct substitution: (2)2+3(2)−5=4+6−5=5(2)^2 + 3(2) - 5 = 4 + 6 - 5 = 5(2)2+3(2)−5=4+6−5=5. Therefore, the limit is 5.
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How to solve limxo2(x2+3x−5)lim_{x o 2} (x^2 + 3x - 5)limxo2​(x2+3x−5)?

  1. Direct substitution: (2)2+3(2)−5=4+6−5=5(2)^2 + 3(2) - 5 = 4 + 6 - 5 = 5(2)2+3(2)−5=4+6−5=5. Therefore, the limit is 5.

How to solve limxo3x2−9x−3lim_{x o 3} \frac{x^2 - 9}{x - 3}limxo3​x−3x2−9​?

  1. Direct substitution yields 0/0. 2. Factor the numerator: x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3)x2−9=(x−3)(x+3). 3. Simplify: (x−3)(x+3)x−3=x+3\frac{(x - 3)(x + 3)}{x - 3} = x + 3x−3(x−3)(x+3)​=x+3. 4. Evaluate the limit: limxo3(x+3)=3+3=6lim_{x o 3} (x + 3) = 3 + 3 = 6limxo3​(x+3)=3+3=6.

How to solve limxo0x+4−2xlim_{x o 0} \frac{\sqrt{x + 4} - 2}{x}limxo0​xx+4​−2​?

  1. Direct substitution yields 0/0. 2. Rationalize the numerator by multiplying by the conjugate: x+4−2xcdotx+4+2x+4+2\frac{\sqrt{x + 4} - 2}{x} cdot \frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2}xx+4​−2​cdotx+4​+2x+4​+2​. 3. Simplify: x+4−4x(x+4+2)=xx(x+4+2)=1x+4+2\frac{x + 4 - 4}{x(\sqrt{x + 4} + 2)} = \frac{x}{x(\sqrt{x + 4} + 2)} = \frac{1}{\sqrt{x + 4} + 2}x(x+4​+2)x+4−4​=x(x+4​+2)x​=x+4​+21​. 4. Evaluate the limit: limxo01x+4+2=14+2=14lim_{x o 0} \frac{1}{\sqrt{x + 4} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}limxo0​x+4​+21​=4​+21​=41​.

How to solve limxo5(2x2−3)lim_{x o 5} (2x^2 - 3)limxo5​(2x2−3)?

  1. Direct substitution: 2(5)^2 - 3 = 2(25) - 3 = 50 - 3 = 47. Therefore, the limit is 47.

How to solve limxo−2x3+8x+2lim_{x o -2} \frac{x^3 + 8}{x + 2}limxo−2​x+2x3+8​?

  1. Direct substitution yields 0/0. 2. Factor the numerator: x3+8=(x+2)(x2−2x+4)x^3 + 8 = (x + 2)(x^2 - 2x + 4)x3+8=(x+2)(x2−2x+4). 3. Simplify: (x+2)(x2−2x+4)x+2=x2−2x+4\frac{(x + 2)(x^2 - 2x + 4)}{x + 2} = x^2 - 2x + 4x+2(x+2)(x2−2x+4)​=x2−2x+4. 4. Evaluate the limit: limxo−2(x2−2x+4)=(−2)2−2(−2)+4=4+4+4=12lim_{x o -2} (x^2 - 2x + 4) = (-2)^2 - 2(-2) + 4 = 4 + 4 + 4 = 12limxo−2​(x2−2x+4)=(−2)2−2(−2)+4=4+4+4=12.

How to solve limxo4x−2x−4lim_{x o 4} \frac{\sqrt{x} - 2}{x - 4}limxo4​x−4x​−2​?

  1. Direct substitution yields 0/0. 2. Rationalize the numerator: x−2x−4cdotx+2x+2\frac{\sqrt{x} - 2}{x - 4} cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2}x−4x​−2​cdotx​+2x​+2​. 3. Simplify: x−4(x−4)(x+2)=1x+2\frac{x - 4}{(x - 4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}(x−4)(x​+2)x−4​=x​+21​. 4. Evaluate the limit: limxo41x+2=14+2=14lim_{x o 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{4}limxo4​x​+21​=4​+21​=41​.

How to solve limxo1x2+4x−5x−1lim_{x o 1} \frac{x^2 + 4x - 5}{x - 1}limxo1​x−1x2+4x−5​?

  1. Direct substitution yields 0/0. 2. Factor the numerator: x2+4x−5=(x−1)(x+5)x^2 + 4x - 5 = (x - 1)(x + 5)x2+4x−5=(x−1)(x+5). 3. Simplify: (x−1)(x+5)x−1=x+5\frac{(x - 1)(x + 5)}{x - 1} = x + 5x−1(x−1)(x+5)​=x+5. 4. Evaluate the limit: limxo1(x+5)=1+5=6lim_{x o 1} (x + 5) = 1 + 5 = 6limxo1​(x+5)=1+5=6.

How to solve limxo0(x+5)2−25xlim_{x o 0} \frac{(x+5)^2 - 25}{x}limxo0​x(x+5)2−25​?

  1. Direct substitution yields 0/0. 2. Expand: x2+10x+25−25x=x2+10xx\frac{x^2+10x+25 - 25}{x} = \frac{x^2+10x}{x}xx2+10x+25−25​=xx2+10x​. 3. Simplify: x(x+10)x=x+10\frac{x(x+10)}{x} = x+10xx(x+10)​=x+10. 4. Evaluate the limit: limx→0(x+10)=0+10=10lim_{x \to 0} (x+10) = 0+10 = 10limx→0​(x+10)=0+10=10.

How to solve limxo2(5x−3x3)lim_{x o 2} (5x - 3x^3)limxo2​(5x−3x3)?

  1. Direct substitution: 5(2) - 3(2)^3 = 10 - 3(8) = 10 - 24 = -14. Therefore, the limit is -14.

How to solve limxo−1x2−1x+1lim_{x o -1} \frac{x^2 - 1}{x + 1}limxo−1​x+1x2−1​?

  1. Direct substitution yields 0/0. 2. Factor the numerator: x2−1=(x−1)(x+1)x^2 - 1 = (x - 1)(x + 1)x2−1=(x−1)(x+1). 3. Simplify: (x−1)(x+1)x+1=x−1\frac{(x - 1)(x + 1)}{x + 1} = x - 1x+1(x−1)(x+1)​=x−1. 4. Evaluate the limit: limxo−1(x−1)=−1−1=−2lim_{x o -1} (x - 1) = -1 - 1 = -2limxo−1​(x−1)=−1−1=−2.

What is the Sum Rule for limits?

limxoc[f(x)+g(x)]=limxocf(x)+limxocg(x)lim_{x o c} [f(x) + g(x)] = lim_{x o c} f(x) + lim_{x o c} g(x)limxoc​[f(x)+g(x)]=limxoc​f(x)+limxoc​g(x)

What is the Difference Rule for limits?

limxoc[f(x)−g(x)]=limxocf(x)−limxocg(x)lim_{x o c} [f(x) - g(x)] = lim_{x o c} f(x) - lim_{x o c} g(x)limxoc​[f(x)−g(x)]=limxoc​f(x)−limxoc​g(x)

What is the Constant Multiple Rule for limits?

limxoc[kcdotf(x)]=kcdotlimxocf(x)lim_{x o c} [k cdot f(x)] = k cdot lim_{x o c} f(x)limxoc​[kcdotf(x)]=kcdotlimxoc​f(x)

What is the Product Rule for limits?

limxoc[f(x)cdotg(x)]=limxocf(x)cdotlimxocg(x)lim_{x o c} [f(x) cdot g(x)] = lim_{x o c} f(x) cdot lim_{x o c} g(x)limxoc​[f(x)cdotg(x)]=limxoc​f(x)cdotlimxoc​g(x)

What is the Quotient Rule for limits?

limxocf(x)g(x)=limxocf(x)limxocg(x)lim_{x o c} \frac{f(x)}{g(x)} = \frac{lim_{x o c} f(x)}{lim_{x o c} g(x)}limxoc​g(x)f(x)​=limxoc​g(x)limxoc​f(x)​, if lim_{x o c} g(x) eq 0

What is the Power Rule for limits?

limxoc[f(x)]n=[limxocf(x)]nlim_{x o c} [f(x)]^n = [lim_{x o c} f(x)]^nlimxoc​[f(x)]n=[limxoc​f(x)]n

What is the Root Rule for limits?

limxocf(x)n=limxocf(x)nlim_{x o c} \sqrt[n]{f(x)} = \sqrt[n]{lim_{x o c} f(x)}limxoc​nf(x)​=nlimxoc​f(x)​

Formula to find limx→cklim_{x \to c} klimx→c​k where k is a constant?

limx→ck=klim_{x \to c} k = klimx→c​k=k

Formula to find limx→cxlim_{x \to c} xlimx→c​x?

limx→cx=clim_{x \to c} x = climx→c​x=c

Formula to find limx→cxnlim_{x \to c} x^nlimx→c​xn?

limx→cxn=cnlim_{x \to c} x^n = c^nlimx→c​xn=cn

Explain the concept of direct substitution for evaluating limits.

First attempt to evaluate a limit by plugging in the value that x approaches. If it results in a defined value, that is the limit.

What should you do if direct substitution results in an indeterminate form?

Use algebraic manipulation (factoring, rationalizing, etc.) or other techniques to simplify the expression before evaluating the limit.

Explain how limit laws help in evaluating complex limits.

Limit laws allow you to break down complex limits into simpler parts by dealing with sums, differences, products, quotients, and powers separately.

Explain the importance of checking the denominator when using the Quotient Rule.

The Quotient Rule can only be applied if the limit of the denominator is not zero. Otherwise, the limit does not exist or requires further analysis.

What happens to the limit when there is no 'x' in the function?

If the function is a constant, the limit is simply that constant value, regardless of what x approaches.

Explain the concept of one-sided limits.

One-sided limits examine the behavior of a function as x approaches a value from the left (x -> c-) or from the right (x -> c+).

How do one-sided limits relate to the existence of a two-sided limit?

For a two-sided limit to exist at a point, both the left-hand limit and the right-hand limit must exist and be equal at that point.

What is the first step in evaluating any limit?

Attempt direct substitution. If it yields a defined value, the limit is found. If it yields an indeterminate form, further techniques are required.

When can you apply the limit laws?

Limit laws can be applied when the individual limits of the functions involved exist. They help simplify complex expressions into manageable components.

Explain the relationship between limits and continuity.

For a function to be continuous at a point, the limit must exist at that point, the function must be defined at that point, and the limit must equal the function's value at that point.