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I need to solve this limit $$\lim_{x\to 5} \frac{5^{x}-x^5}{x-5}$$ without using l'Hospital.

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    $\begingroup$ How about: $$ \frac{5^{x}-x^5}{x-5}$$=$$ \frac{5^{x}-5^5}{x-5}+ \frac{5^{5}-x^5}{x-5}$$ $\endgroup$ – Juanito May 14 '17 at 18:34
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Hint:

\begin{align*} \lim_{x\to 5} \frac{5^{x}-x^5}{x-5}&=\lim_{x\to 5}\left(\frac{5^x-5^5}{x-5}-\frac{x^5-5^5}{x-5}\right)\\[4pt] &=\lim_{x\to 5}\frac{5^x-5^5}{x-5}-\lim_{x\to 5}\frac{x^5-5^5}{x-5} \end{align*} These two limits are the derivatives of $5^x$ and $x^5$ at $x=5$.

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$$ \begin{aligned} \lim_{x\to 5} \frac{5^{x}-x^5}{x-5} & =\lim _{t\to \:0}\:\frac{5^{\left(t+5\right)}-\left(t+5\right)^5}{\left(t+5\right)-5} \\& = \lim _{t\to \:0}\:\left(\frac{5^{t+5}-t^5-25t^4-250t^3-1250t^2-3125t-3125}{t}\right) \\& = \lim _{t\to \:0}\:\left(\frac{\left(3125+3125\ln \:\left(5\right)t+o\left(t\right)\right)-t^5-25t^4-250t^3-1250t^2-3125t-3125}{t}\right) \\& = \lim _{t\to \:0}\:\left(-t^4-25t^3-250t^2-1250t-3125+3125\ln \:\left(5\right)\right) \\& = \color{red}{3125\left(\ln \left(5\right)-1\right)} \end{aligned} $$

Solved with Taylor expansion

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HINT:

$$\lim_{x\to5}\dfrac{5^x-5^5-(x^5-5^5)}{x-5}$$

$$=5^5\lim_{x\to5}\dfrac{5^{x-5}-1}{x-5}-\lim_{x\to5}\dfrac{x^5-5^5}{x-5}$$

For the first part, $$\lim_{h\to}\dfrac{a^h-1}h=\ln a$$

For the second as $x-5\to0,x-5\ne0,$ $$\dfrac{x^5-5^5}{x-5}=\sum_{r=0}^4x^{4-r}5^r$$

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