How would I solve for something like this??
$$\lim_{x\to 5} \frac{2^x - 32}{x-5}$$
using the definition of derivatives.
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3Short Answer: $f(x)=2^x$ and the limit is the same as $f'(5)$. – Noble Mushtak Feb 13 '16 at 00:59
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2Shorter answer: by trying something, to start with. – Clement C. Feb 13 '16 at 01:00
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3Especially since it's exactly the same idea as in your previous question. – Clement C. Feb 13 '16 at 01:02
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2Hint: The derivative of $f(x)$ at $a$ is $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. Compare. – André Nicolas Feb 13 '16 at 01:02
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I thought it might be instructive to show a way forward that doesn't rely on calculus. Note we can write
$$\begin{align} \frac{2^x-32}{x-5}&=32\frac{2^{x-5}-1}{x-5}\\\\ &=32\left(\frac{e^{\log(2)\,(x-5)}-1}{x-5}\right) \end{align}$$
In THIS ANSWER, I showed using on the limit definition of the exponential function along with Bernoulli's Inequality the exponential function satisfies the inequalities
$$\frac{1}{1-x}\ge e^x\ge 1+x$$
for $x<1$. Therefore, we can assert that
$$\frac{\log(2)\,(x-5)}{x-5}\le \frac{e^{\log(2)\,(x-5)}-1}{x-5}\le \frac{\frac{\log(2)\,(x-5)}{1-\log(2)\,(x-5)}}{x-5}$$
whereupon applying the squeeze theorem gives
$$\lim_{x\to 5}\frac{2^x-32}{x-5}=32\log(2)$$
And we are done!
Mark Viola
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