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Find the limit. \lim_{ x \rightarrow 2}\frac{1}{1+e^{1/x-2}}

Question:

Find the limit.

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

Exponential Function in Denominator:

If the value of the limit is not equal to {eq}\frac{0}{0}, \frac{\infty }{\infty }, \infty ^{0} {/eq} etc. that means there is no indeterminate form of the limit so we can compute the value of the given limit by direct substitution method.

If the exponent of the exponential function is infinity then the value of this function is equal to infinity.

Answer and Explanation:

The quotient limit expression is:

{eq}\displaystyle L= \lim_{ x \rightarrow 2}\frac{1}{1+e^{(\frac{1}{x - 2})}} {/eq}

Substituting the value of the variable {eq}x {/eq} in the above limit expression, we get:

{eq}\begin{align*} \displaystyle L&=\frac{1}{1+e^{(\frac{1}{2 - 2})}}\\ &=\frac{1}{1+e^{(\frac{1}{0})}}\\ &=\frac{1}{1+e^{\infty }}&\because \frac{1}{0}=\infty \\ &=\frac{1}{1+\infty}\\ &=\frac{1}{\infty}\\ &=\frac{1}{\frac{1}{0}}\\ &=0 \end{align*} {/eq}


Learn more about this topic:

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How to Determine the Limits of Functions

from Math 104: Calculus

Chapter 6 / Lesson 4
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