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

Question

Scarlett Allison · Accepted Answer

# f'(2) = ln 16 \ \ \ # QED Given the context of the question, we can see that this limit is the derivative of the function #f(x)=2^x# when #x=2#, Using the limit definition of the derivative, we calculate this using: # f'(2) = lim_(x rarr2) ( f(x)-f(2) ) / (x-2) # # \ \ \ \ \ \ \ \ = lim_(x rarr2) ( 2^x-2^2 ) / (x-2) # # \ \ \ \ \ \ \ \ = lim_(x rarr2) ( 2^x-4 ) / (x-2) # Let us perform a substitution, where: # u = x-2 # Clearly as #x rarr 2 => u rarr 0#, and so we can rewrite the limits as: # f'(2) = lim_(u rarr 0) ( 2^(u+2)-4 ) / (u) # # \ \ \ \ \ \ \ \ = lim_(u rarr 0) ( 2^2 2^u-4 ) / (u) # # \ \ \ \ \ \ \ \ = 4 \ lim_(u rarr 0) ( 2^u-1 ) / (u) # Now we require some additional knowledge of logarithms and specific calculus limits, which provides the value of the limit # lim_(u rarr 0) ( a^u-1 ) / (u) = ln a# Where #ln# denotes the Natural Logarithm, ie a logarithm to base #e#, where #e# is Euler's Number. Using this result we get: # f'(2) = 4 \ ln2 # # \ \ \ \ \ \ \ \ = ln 2^4 # # \ \ \ \ \ \ \ \ = ln 16 \ \ \ # QED

Scarlett Allison · Answer

undefined 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 $.