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Often, when faced with an indeterminate limit, there is a much simpler way to solve the problem compared with applying L'Hopital's rule. With that in mind, compute the following. Use inf to represent positive infinity, -inf to represent negative infinity, and dne to represent the limit not existing in a different way. (a) lim 8个8 4x8+2x+175 7x8+1 4/7 ez (b) lim * (d) lim 1000x5+ sin(x) log(5) 80+x x2 0 = Inf (d) lim (√x8 +1-√√√x² - 1) = \\lim _{\rightarrow -\infty }(x^{4}1+x8 x-x : Consider the function f(x): lim f(x) = 1 = exte-x ex-e-x. Use asymptotic reasoning to determine lim ƒ(x). x→∞ x→∞ On the other hand, the above limit has the indeterminate form ∞, so we might consider using L'Hopital's rule. What limit are you left with after applying the rule once? (Enter an expression involving x.) lim f(x) = lim (e^(x)-e^(-x))/(e^(x)+e^(-x) x→∞ 81x The limit you just obtained still has the indeterminate form the rule again? ∞ so we might consider applying L'Hopital's rule again. What limit are you left with after applying lim f(x) = lim (e^(x)+e^(-x))/(e^(x)-e^(-x) x+x 8个8 Was L'Hopital's rule helpful in computing the limit? No, because we just got back the original function

Often, when faced with an indeterminate limit, there is a much simpler way to solve the problem compared with applying L'Hopital's rule. With that in mind, compute the following. Use inf to represent positive infinity, -inf to represent negative infinity, and dne to represent the limit not existing in a different way. (a) lim 8个8 4x8+2x+175 7x8+1 4/7 ez (b) lim * (d) lim 1000x5+ sin(x) log(5) 80+x x2 0 = Inf (d) lim (√x8 +1-√√√x² - 1) = \\lim _{\rightarrow -\infty }(x^{4}1+x8 x-x : Consider the function f(x): lim f(x) = 1 = exte-x ex-e-x. Use asymptotic reasoning to determine lim ƒ(x). x→∞ x→∞ On the other hand, the above limit has the indeterminate form ∞, so we might consider using L'Hopital's rule. What limit are you left with after applying the rule once? (Enter an expression involving x.) lim f(x) = lim (e^(x)-e^(-x))/(e^(x)+e^(-x) x→∞ 81x The limit you just obtained still has the indeterminate form the rule again? ∞ so we might consider applying L'Hopital's rule again. What limit are you left with after applying lim f(x) = lim (e^(x)+e^(-x))/(e^(x)-e^(-x) x+x 8个8 Was L'Hopital's rule helpful in computing the limit? No, because we just got back the original function

Intermediate Algebra
Intermediate Algebra
19th Edition
ISBN:9780998625720
Author:Lynn Marecek
Publisher:Lynn Marecek
Chapter8: Roots And Radicals
Section8.7: Use Radicals In Functions
Problem 408E: Explain why the process of finding the domain of a radical function with an even index is different...
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part C and D in each of the questions are wrong, Also i just need the answer

Often, when faced with an indeterminate limit, there is a much simpler way to solve the problem compared with applying L'Hopital's rule. With that in mind, compute the following. Use inf to represent positive infinity, -inf to represent negative infinity, and dne to represent the limit not existing in a different way. (a) lim 8个8 4x8+2x+175 7x8+1 4/7 ez (b) lim * (d) lim 1000x5+ sin(x) log(5) 80+x x2 0 = Inf (d) lim (√x8 +1-√√√x² - 1) = \\lim _{\rightarrow -\infty }(x^{4}1+x8 x-x :
Transcribed Image Text:Often, when faced with an indeterminate limit, there is a much simpler way to solve the problem compared with applying L'Hopital's rule. With that in mind, compute the following. Use inf to represent positive infinity, -inf to represent negative infinity, and dne to represent the limit not existing in a different way. (a) lim 8个8 4x8+2x+175 7x8+1 4/7 ez (b) lim * (d) lim 1000x5+ sin(x) log(5) 80+x x2 0 = Inf (d) lim (√x8 +1-√√√x² - 1) = \\lim _{\rightarrow -\infty }(x^{4}1+x8 x-x :
Consider the function f(x): lim f(x) = 1 = exte-x ex-e-x. Use asymptotic reasoning to determine lim ƒ(x). x→∞ x→∞ On the other hand, the above limit has the indeterminate form ∞, so we might consider using L'Hopital's rule. What limit are you left with after applying the rule once? (Enter an expression involving x.) lim f(x) = lim (e^(x)-e^(-x))/(e^(x)+e^(-x) x→∞ 81x The limit you just obtained still has the indeterminate form the rule again? ∞ so we might consider applying L'Hopital's rule again. What limit are you left with after applying lim f(x) = lim (e^(x)+e^(-x))/(e^(x)-e^(-x) x+x 8个8 Was L'Hopital's rule helpful in computing the limit? No, because we just got back the original function
Transcribed Image Text:Consider the function f(x): lim f(x) = 1 = exte-x ex-e-x. Use asymptotic reasoning to determine lim ƒ(x). x→∞ x→∞ On the other hand, the above limit has the indeterminate form ∞, so we might consider using L'Hopital's rule. What limit are you left with after applying the rule once? (Enter an expression involving x.) lim f(x) = lim (e^(x)-e^(-x))/(e^(x)+e^(-x) x→∞ 81x The limit you just obtained still has the indeterminate form the rule again? ∞ so we might consider applying L'Hopital's rule again. What limit are you left with after applying lim f(x) = lim (e^(x)+e^(-x))/(e^(x)-e^(-x) x+x 8个8 Was L'Hopital's rule helpful in computing the limit? No, because we just got back the original function
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