Evaluating $\lim_{x\to 1}\frac{8^x-8}{x-1}$ without using L'Hopital rule

Question

Evaluate without using L'Hopital rule $$\lim_{x\to 1}\frac{8^x-8}{x-1}$$

Help me with this limit.

Answer 1

Use the substitution $t = x - 1$,

$$ \lim_{{x \to 1}} \frac{{8^x - 8}}{{x - 1}} $$

with $x = t + 1$.

Substitute $t$ into the expression:

$$ \lim_{t \to 0} \frac{{8^{t + 1} - 8}}{{t}}$$

and we have $$\lim_{t \to 0} \frac{{8(8^t - 1)}}{{t}} $$

Remember that

$$\lim_{t \to 0} \frac{a^t-1}{{t}}=\log a. $$

$\therefore, \ \lim\cdots =8\log 8$.

Answer 2

Hint: Use the Taylor series of the function $f(x) = 8^{x}$ at $x = 1$.