# Prove that #lim_(x rarr2) ( 2^x-4 ) / (x-2) =ln16#?

# f'(2) = ln 16 \ \ \ # QED

Let us perform a substitution, where:

Now we require some additional knowledge of logarithms and specific calculus limits, which provides the value of the limit

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To prove ( \lim_{x \to 2} \frac{2^x - 4}{x - 2} = \ln 16 ), we can use L'Hôpital's Rule or evaluate the limit directly by recognizing the form ( \frac{0}{0} ) as ( x ) approaches 2.

Using L'Hôpital's Rule, we take the derivative of the numerator and denominator with respect to ( x ), then evaluate the limit again:

[ \lim_{x \to 2} \frac{2^x - 4}{x - 2} = \lim_{x \to 2} \frac{\frac{d}{dx}(2^x - 4)}{\frac{d}{dx}(x - 2)} ]

[ = \lim_{x \to 2} \frac{\ln(2) \cdot 2^x}{1} = \ln(2) \cdot 2^2 = \ln(4) = \ln(16) ]

Alternatively, we can factor ( 2^x - 4 ) as ( 2^2(2^{x - 2} - 1) ):

[ \lim_{x \to 2} \frac{2^x - 4}{x - 2} = \lim_{x \to 2} \frac{2^2(2^{x - 2} - 1)}{x - 2} = 2^2 \lim_{x \to 2} \frac{2^{x - 2} - 1}{x - 2} ]

Recognizing the form ( \frac{0}{0} ), we can evaluate the limit directly:

[ = 2^2 \cdot \ln(2) = \ln(4) = \ln(16) ]

Therefore, ( \lim_{x \to 2} \frac{2^x - 4}{x - 2} = \ln 16 ).

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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.

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.

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.

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