Limit of log functions [closed]

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I need help solving this problem.

$$\displaystyle \lim _{x\to 0}\frac{\log\left(1+7x\right)}{5x}$$

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7 Answers

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hint: $Q = \dfrac{7}{5}\cdot\dfrac{\log(1+7x)}{7x}$

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$\lim_{x \to 0} \frac{\log(1+7x)}{5x}=\frac{0}{0}.$

Since the limit is of an indeterminate type, we can apply L'Hospitals rule. That is, just differentiate the top and differentiate the bottom and then reevaluate the limit.

$\lim_{x \to 0} \frac{\log(1+7x)}{5x}=\frac{0}{0} \;\;\; \Rightarrow \;\;\; \lim_{x \to 0} {\frac{\frac{1}{1+7x} \cdot 7}{5}} = \frac{7}{5}$

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$$\lim _{x\to 0}\frac{\log\left(1+7x\right)}{5x}=\lim_{x\to 0}\frac{7}{5}\frac{1}{7x}\log{(1+7x)}=$$ $$=\frac{7}{5}\log{\left(\lim_{x\to 0}(1+7x)^{\frac{1}{7x}}\right)}=\frac{7}{5}\log e=\frac{7}{5}$$

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Hint

You can also use the fact that, if $y<<1$, then $\log(1+y)\approx y$. Replace $y$ by $7x$ and you will get the answer.

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Use the definition of log's derivative and set $7x = t$: $$ \frac{7}{5} \lim_{t \to 0} \lim_{v \to 1} \frac{\log (v + t) - \log v }{t} = \frac{7}{5} \lim_{v \to 1} \lim_{t \to 0}\frac{\log (v + t) - \log v }{t} = \frac{7}{5} \lim_{v \to 1} \frac{1}{v} = \frac{7}{5} $$

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I think you can solve this by using the following formula:
$$ \lim _{x\to 0}\frac{\log(1+x)}{x}=1$$

So just divide and multiply by $7$, you will get the answer as $7/5$

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We can expand $\log(1+x)$ for $|x|\lt1$ as:

$$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\dots$$

Therefore:

$$\lim_\limits{x\to0}\frac{\log(1+7x)}{5x}$$ $$=\lim_\limits{x\to0}\frac{7x-\frac{49x^2}2+\dots}{5x}$$ $$=\frac75$$

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