Chapter 7: Q. 1 TB (page 638)

Explain why $\lim_{k \to \infty} k^{\frac{1}{k}}$ has an indeterminate form of the type $\infty^{0} .$ Then show that this limit equals $1 .$

Short Answer

Expert verified

Limit $\lim_{k \to \infty} k^{\frac{1}{k}}$ has an indeterminate form of the type $\infty^{0}$ by substituting the limit value for k.

$\lim_{k \to \infty} k^{\frac{1}{k}} = \infty^{\frac{1}{\infty}} = \infty^{0}$

The value of the limit is $1$ as follows.

$\lim_{k \to \infty} k^{\frac{1}{k}} = \lim_{k \to \infty} e^{\ln (k^{\frac{1}{k}})} = \lim_{k \to \infty} e^{\frac{1}{k} \ln k} = e^{0} = 1$

Step by step solution

Step 1. Given information.

Limit $\lim_{k \to \infty} k^{\frac{1}{k}}$ has an indeterminate form of the type $\infty^{0} .$

Step 2. Verifying the indeterminate form of the type ∞0.

Determine the value of the given limit by substituting the limit value for k.

$\lim_{k \to \infty} k^{\frac{1}{k}} = \infty^{\frac{1}{\infty}} = \infty^{0}$

so $\lim_{k \to \infty} k^{\frac{1}{k}}$ has an indeterminate form of the type $\infty^{0} .$

Step 2. Value of limit.

Determine the value of the limit by using the logarithm property $e^{\ln x} = x .$

role="math" localid="1649194572093" $\lim_{k \to \infty} k^{\frac{1}{k}} = \lim_{k \to \infty} e^{\ln (k^{\frac{1}{k}})} = \lim_{k \to \infty} e^{\frac{1}{k} \ln k} = e^{\frac{1}{\infty} \ln \infty} = e^{0} = 1$

So $\lim_{k \to \infty} k^{\frac{1}{k}} = = 1 .$

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Explain why $\lim_{k \to \infty} k^{\frac{1}{k}}$ has an indeterminate form of the type $\infty^{0} .$ Then show that this limit equals $1 .$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Verifying the indeterminate form of the type ∞0.

Step 2. Value of limit.

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Explain why limk→∞k1khas an indeterminate form of the type ∞0.Then show that this limit equals1.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Verifying the indeterminate form of the type ∞0.

Step 2. Value of limit.

One App. One Place for Learning.

Most popular questions from this chapter

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Explain why $\lim_{k \to \infty} k^{\frac{1}{k}}$ has an indeterminate form of the type $\infty^{0} .$ Then show that this limit equals $1 .$