Question

Lim x→0 (log 1+8x)/x

Answer 1

$\underline{\textsf{To find:}}$

$\textsf{Value of}$

$\mathsf{\displaystyle\lim_{x\to\;0}\;\dfrac{log(1+8x)}{x}}$

$\underline{\textsf{Solution:}}$

$\mathsf{Consider,}$

$\mathsf{\displaystyle\lim_{x\to\;0}\;\dfrac{log(1+8x)}{x}}$

$\mathsf{when\;applying\;limit,\;we\;get\;\dfrac{0}{0}\;form}$

$\mathsf{Applying\;L\;Hopital's\;rule}$

$\mathsf{\displaystyle\lim_{x\to\;0}\;\dfrac{log(1+8x)}{x}=\displaystyle\lim_{x\to\;0}\;\dfrac{\left(\dfrac{1}{1+8x}\right)8}{1}}$

$\mathsf{=\dfrac{\left(\dfrac{1}{1+8(0)}\right)8}{1}}$

$\mathsf{=8}$

$\therefore\boxed{\mathsf{\displaystyle\lim_{x\to\;0}\;\dfrac{log(1+8x)}{x}=8}}$

$\underline{\textsf{Find more:}}$

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Answer 2

SOLUTION

TO DETERMINE

$\displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + 8x)}{x}$

FORMULA TO BE IMPLEMENTED

We are aware of the formula on limit that

$\displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + x)}{x} = 1$

EVALUATION

Here the given limit

$= \: \displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + 8x)}{x}$

$= \: \displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + 8x)}{8x} \: \times 8$

$= \: 8 \times \displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + 8x)}{8x}$

Let u = 8x

Then as x → 0 we have u → 0

Then the given limit becomes

$\: \displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + 8x)}{x}$

$= \: 8 \times \displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + 8x)}{8x}$

$= \: 8 \times \displaystyle \lim_{u \to 0} \: \frac{ \log \: (1 + u)}{u}$

$= 8 \times 1 \: \: \bigg( \: \because \: \displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + x)}{x} = 1 \bigg)$

$= 8$

FINAL ANSWER

$\boxed{ \: \: \displaystyle \lim_{x \to 0} \: \frac{ \log \: (1 + 8x)}{x} \: = \: 8 \: \: \: }$

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