Evaluate the following limits. If needed, enter infinity for ∞ and -infinity for −∞ a) limx→92+(29x9−2x) b) limx→92−(29x9−2x)

a) We have to evaluate the folllowing limits: limx→92+(29x9−2x) ∴limx→92+cf(x)=climx→92+f(x) Here c=29 and f(x)=x9−2x ⇒limx→92+29x9−2x=29⋅limx→92+x9−2x The function decreasses without a bound ⇒limx→92+x9−2x=−∞ Therefore limx→92+29x9−2x=−∞

b) We have to evaluate the folllowing limits: limx→92−(29x9−2x) ∴limx→92−cf(x)=climx→92−f(x) Here c=29 and f(x)=x9−2x ⇒limx→92−29x9−2x=29⋅limx→92−x9−2x The function increasses without a bound ⇒limx→92−x9−2x=∞ Therefore limx→92−29x9−2x=∞

a) We have to evaluate the folllowing limits: limx→92+(29x9−2x) ∴limx→92+cf(x)=climx→92+f(x) Here c=29 and f(x)=x9−2x ⇒limx→92+29x9−2x=29⋅limx→92+x9−2x The function decreasses without a bound ⇒limx→92+x9−2x=−∞ Therefore limx→92+29x9−2x=−∞

b) We have to evaluate the folllowing limits: limx→92−(29x9−2x) ∴limx→92−cf(x)=climx→92−f(x) Here c=29 and f(x)=x9−2x ⇒limx→92−29x9−2x=29⋅limx→92−x9−2x The function increasses without a bound ⇒limx→92−x9−2x=∞ Therefore limx→92−29x9−2x=∞

"Evaluate the following limits. If needed, enter infinity..." was answered by students like you

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Answered question

2021-11-19

Evaluate the following limits. If needed, enter infinity for

\infty

and -infinity for

- \infty

lim_{x \to {\frac{9}{2}}^{+}} (\frac{29 x}{9 - 2 x})

lim_{x \to {\frac{9}{2}}^{-}} (\frac{29 x}{9 - 2 x})

Answer & Explanation

Aretha Frazier

Beginner2021-11-20Added 16 answers

a) We have to evaluate the folllowing limits:

lim_{x \to {\frac{9}{2}}^{+}} (\frac{29 x}{9 - 2 x})

∴ lim_{x \to {\frac{9}{2}}^{+}} c f (x) = c lim_{x \to {\frac{9}{2}}^{+}} f (x)

Here c=29 and

f (x) = \frac{x}{9 - 2 x}

\Rightarrow lim_{x \to {\frac{9}{2}}^{+}} \frac{29 x}{9 - 2 x} = 29 \cdot lim_{x \to {\frac{9}{2}}^{+}} \frac{x}{9 - 2 x}

The function decreasses without a bound

\Rightarrow lim_{x \to {\frac{9}{2}}^{+}} \frac{x}{9 - 2 x} = - \infty

Therefore

lim_{x \to {\frac{9}{2}}^{+}} \frac{29 x}{9 - 2 x} = - \infty

Poul1963

Beginner2021-11-21Added 15 answers

b) We have to evaluate the folllowing limits:

lim_{x \to {\frac{9}{2}}^{-}} (\frac{29 x}{9 - 2 x})

∴ lim_{x \to {\frac{9}{2}}^{-}} c f (x) = c lim_{x \to {\frac{9}{2}}^{-}} f (x)

Here c=29 and

f (x) = \frac{x}{9 - 2 x}

\Rightarrow lim_{x \to {\frac{9}{2}}^{-}} \frac{29 x}{9 - 2 x} = 29 \cdot lim_{x \to {\frac{9}{2}}^{-}} \frac{x}{9 - 2 x}

The function increasses without a bound

\Rightarrow lim_{x \to {\frac{9}{2}}^{-}} \frac{x}{9 - 2 x} = \infty

Therefore

lim_{x \to {\frac{9}{2}}^{-}} \frac{29 x}{9 - 2 x} = \infty

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