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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=472(5)^2 - 3 = 2(25) - 3 = 50 - 3 = 472(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=−145(2) - 3(2)^3 = 10 - 3(8) = 10 - 24 = -145(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 limxocg(x)eq0lim_{x o c} g(x) eq 0limxoc​g(x)eq0

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

What is a limit?

The value that a function approaches as the input approaches some value.

What is direct substitution in the context of limits?

Evaluating a limit by plugging in the value that x is approaching into the function.

What is an indeterminate form?

An expression whose value cannot be determined, such as 0/0.

Define the Sum Rule for limits.

The limit of a sum is the sum of the 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)

Define the Difference Rule for limits.

The limit of a difference is the difference of the 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)

Define the Constant Multiple Rule for limits.

The limit of a constant times a function is the constant times the limit of the function: 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)

Define the Product Rule for limits.

The limit of a product is the product of the 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)

Define the Quotient Rule for limits.

The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero: 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 limxocg(x)eq0lim_{x o c} g(x) eq 0limxoc​g(x)eq0

Define the Power Rule for limits.

The limit of a function raised to a power is the limit of the function raised to that power: 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

Define the Root Rule for limits.

The limit of a root of a function is the root of the limit of the function: 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)​