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How do you find the limit of # (2^x-3^-x)/(2^x+3^-x)# as x approaches infinity?

Answer 1

Samantha Adkison

#L = lim_{x \to \infty} (2^x-3^-x)/(2^x+3^-x) #

#= lim_{x \to \infty} (2^x-3^-x)/(2^x+3^-x) * (2^{-x})/(2^{-x})#

#= lim_{x \to \infty} (1-3^{-x}*2^{-x})/(1+3^{-x}*2^{-x}) #

#= lim_{x \to \infty} (1 - 6^{-x} )/(1+6^{-x}) #

#lim_{x \to infty} 6^{-x} = 0 \implies L = 1#

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Answer 2

Paisley Johnson

To find the limit of (2^x - 3^-x)/(2^x + 3^-x) as x approaches infinity, we can simplify the expression by multiplying both the numerator and denominator by 3^x. This gives us (2^x * 3^x - 1)/(2^x * 3^x + 1).

Next, we can divide every term in the expression by 3^x. This results in (2^x * (3^x / 3^x) - 1/(3^x))/(2^x * (3^x / 3^x) + 1/(3^x)).

Simplifying further, we have (2^x * 1 - 1/(3^x))/(2^x * 1 + 1/(3^x)).

As x approaches infinity, 1/(3^x) approaches 0. Therefore, the expression becomes (2^x * 1 - 0)/(2^x * 1 + 0), which simplifies to 2^x/2^x, or 1.

Thus, the limit of (2^x - 3^-x)/(2^x + 3^-x) as x approaches infinity is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.