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Find the limit. \lim_{x \rightarrow 0}\frac{x(2^{x})}{(2^{x})-1))}

Question:

Find the limit.

{eq}\displaystyle \lim_{x \rightarrow 0}\frac{x(2^{x})}{2^{x}-1} {/eq}

Limits:

We have a function that has linear and exponential terms. We will apply the L hospital's rule to evaluate the limit. We will differentiate the numerator and the denominator and then apply the limits.

Answer and Explanation:

$$\lim_{x \rightarrow 0}\frac{x(2^{x})}{2^{x}-1}\\ $$

We will apply the L Hospital's rule as the function is of the form:

$$\frac{0}{0}\\ $$

We will differentiate the numertaor and the denominator:

$$=\lim _{x\to \:0}\left(\frac{2^x+\ln \left(2\right)\cdot \:2^xx}{\ln \left(2\right)\cdot \:2^x}\right)\\ =\lim _{x\to \:0}\left(\frac{1+\ln \left(2\right)x}{\ln \left(2\right)}\right)\\ $$

Applying the limits we get:

$$\frac{1}{\ln 2} $$


Learn more about this topic:

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Understanding the Properties of Limits

from Math 104: Calculus

Chapter 6 / Lesson 5
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