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Doesn't seem to be difficult, but still can't get it.

$\displaystyle\lim_{x\rightarrow 0}\frac{10^{-x}-1}{x}=\ln10$

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The answer depends on what you have covered so far in your course. –  Aryabhata Feb 15 '12 at 7:23
    
Think about derivative of $f(x) = 10^{-x}$ at $x = 0$. –  Sammy Black Feb 15 '12 at 7:26

2 Answers 2

up vote 4 down vote accepted

There are a couple of standard approaches. One of them is to rewrite our expression as $$\frac{10^{-x} -10^{-0}}{x-0}.$$ Then you may recognize that our limit is simply the derivative of $10^{-x}$ at $x=0$.

What is that derivative? Since $10=e^{\ln(10)}$, we have that $10^{-x}=e^{-x\ln(10)}$. Differentiate using the Chain rule, and evaluate the derivative at $x=0$. You should get $-\ln(10)$. Please note that the limit is not $\ln(10)$. This is clear without calculation, because for positive $x$, the top is negative, and for negative $x$ the top is positive. So the ratio is always negative when it is defined, and hence the limit, if it exists, must be $\le 0$.

Here is another approach, which is from the point of view of the computations you need to do essentially the same as the previous one. We can use L'Hospital's Rule, since both top and bottom approach $0$ as $x$ approaches $0$.

Remark: If we want to find the above limit directly from the definition of derivative, things get more complicated. Presumably $10^u$ was defined as being $e^{u\ln(10)}$. Note that $$\frac{e^{-x\ln 10}-1}{x}=-\ln (10)\frac{e^{-x\ln 10}-1}{-x\ln(10)}.$$ Now make the substitution $u=-x\ln(10)$. As $x\to 0$, we have $u\to 0$, so our limit is $$-\ln10) \lim_{u\to 0}\frac{e^u-1}{u}$$ if this last limit exists.

The details of this task depend on precisely how the function $e^u$ is defined. One definition I have seen for $e$ is that $e$ is the number such that the above limit is $1$. If we adopt that definition, there is nothing to prove!

If we define $e^u$ as $\lim_{n\to\infty} \left(1+\frac{u}{n}\right)^n$ (after having shown that the limit exists), then the argument depends on making estimates of $(1+u/n)^n$ for $u$ close to $0$, using the Binomial Theorem.

If we first define $\ln(x)$ as the area of the region below $y=1/t$, from $t=1$ to $t=x$, and then define the exponential function as the inverse of the logarithm, then finding the limit we want again depends on estimates, unless, as usual, we first prove a differentiability result for inverse functions.

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thank you! but possibly there is also another way using the limit $\displaystyle\lim_{x->0}(1+x)^{\frac{1}{x}}$, because the task is to find the derivative $10^{-x}$ per definition, that is $\displaystyle\lim_{\Delta{x}\rightarrow0}\frac{10^{-x-\Delta{x}}-10^{-x}}{\Delt‌​a{x}}=10^{-x}\cdot\lim_{\Delta{x}\rightarrow0}\frac{10^{-\Delta{x}}-1}{\Delta{x}}‌​$ –  haemhweg Feb 15 '12 at 7:44

It's worth noticing that we can define the natural logarithm as

$$\log x = \lim_{h \to 0} \frac{x^h-1}{h}$$

So in your case you have

$$ \lim_{h \to 0} \frac{10^{-h}-1}{h}= \lim_{h \to 0} \frac{\frac{1}{10}^{h}-1}{h}=-\log10$$

This result holds because we have that

$$ \lim_{h \to 0} \frac{x^h-1}{h} =\frac{0}{0}$$

So we can apply L'Hôpitals rule, differentiating with respect to $h$ to get

$$ \lim_{h \to 0} \frac{x^h-1}{h} =\lim_{h \to 0} x^h \log x = \log x $$

Obivously this is done by knowing how to handle the derivative of an exponential function with arbitrary base $x$, so you could've also solved you problem by noticing the expression is a derivative, as other answers/comments suggest.

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