Limits
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Question 1 of 6
1. Question
Evaluate`lim_(x->1)(x+1)`- `lim_(x->1)(x+1)=` (2)
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A limit is the value that a function approachesSubstitute the limit to the variable `x``lim_(x->1)(x+1)` `=` `1+1` `=` `2` This means that the limit of `x` as it approaches `1` is `2`. -
Question 2 of 6
2. Question
Evaluate`lim_(x->2)(x^2-4)/(x-2)`- `lim_(x->2)(x^2-4)/(x-2)=` (4)
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A limit is the value that a function approachesFirst, simplify the equation`(x^2-4)/(x-2)` `=` `((x-2)(x+2))/(x-2)` Factorize `=` `x+2` `(x-2)/(x-2)=1` Substitute the limit to the variable `x``lim_(x->2)(x+2)` `=` `2+2` `=` `4` This means that the limit of `x` as it approaches `2` is `4`. -
Question 3 of 6
3. Question
Evaluate`lim_(x->-2)(2x+4)/(x+2)`- `lim_(x->-2)(2x+4)/(x+2)=` (2)
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A limit is the value that a function approachesSimplify the equation`(2x+4)/(x+2)` `=` `(2(x+2))/(x+2)` Factorize `=` `2` `(x+2)/(x+2)=1` This means that the limit of `x` as it approaches `-2` is `2`. -
Question 4 of 6
4. Question
Evaluate`lim_(x->3)(x^2-9)/(x-3)`- `lim_(x->3)(x^2-9)/(x-3)=` (6)
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A limit is the value that a function approachesFirst, simplify the equation`(x^2-9)/(x-3)` `=` `((x-3)(x+3))/(x-3)` Factorize `=` `x+3` `(x-3)/(x-3)=1` Substitute the limit to the variable `x``lim_(x->3)(x+3)` `=` `3+3` `=` `6` This means that the limit of `x` as it approaches `3` is `6`. -
Question 5 of 6
5. Question
Evaluate`lim_(x →∞)(2x^2-5x+7)/(3x^2-4x+2)`Write your answer in the form of “a/b”- `lim_(x →∞)(2x^2-5x+7)/(3x^2-4x+2)=` (2/3)
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A limit is the value that a function approachesFirst, divide the whole expression by the highest power, which is `x^2``(2x^2-5x+7)/(3x^2-4x+2)``divide x^2` `=` $$\frac{\frac{2x^2}{\color{#CC0000}{x^2}}-\frac{5x}{\color{#CC0000}{x^2}}+\frac{7}{\color{#CC0000}{x^2}}}{\frac{3x^2}{\color{#CC0000}{x^2}}-\frac{4x}{\color{#CC0000}{x^2}}+\frac{2}{\color{#CC0000}{x^2}}}$$ `=` $$\frac{2-\frac{5}{x}+\frac{7}{x^2}}{3-\frac{4}{x}+\frac{2}{x^2}}$$ Evaluate Since the limit is infinite, the denominator `x` will keep on increasing and the fraction will keep getting smallerHence, the fractions will be approaching `0``5/x` `->` `0` `7/(x^2)` `->` `0` `4/x` `->` `0` `2/(x^2)` `->` `0` Substitute these values to the equation$$\frac{2-0+0}{3-0+0}$$ `=` $$\frac{2}{3}$$ This means that the limit of `x` as it approaches `∞` is `2/3`. -
Question 6 of 6
6. Question
Evaluate`lim_(x →∞)(x^2-2x+1)/(3x^2-2)`Write your answer in the form of “a/b”- `lim_(x →∞)(x^2-2x+1)/(3x^2-2)=` (1/3)
Hint
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A limit is the value that a function approachesFirst, divide the whole expression by the highest power, which is `x^2``(2x^2-5x+7)/(3x^2-4x+2)``divide x^2` `=` $$\frac{\frac{x^2}{\color{#CC0000}{x^2}}-\frac{2x}{\color{#CC0000}{x^2}}+\frac{1}{\color{#CC0000}{x^2}}}{\frac{3x^2}{\color{#CC0000}{x^2}}-\frac{2}{\color{#CC0000}{x^2}}}$$ `=` $$\frac{1-\frac{2}{x}+\frac{1}{x^2}}{3-\frac{2}{x^2}}$$ Evaluate Since the limit is infinite, the denominator `x` will keep on increasing and the fraction will keep getting smallerHence, the fractions will be approaching `0``2/x` `->` `0` `1/(x^2)` `->` `0` `2/(x^2)` `->` `0` Substitute these values to the equation$$\frac{1-0+0}{3-0}$$ `=` $$\frac{1}{3}$$ This means that the limit of `x` as it approaches `∞` is `1/3`.