4
$\begingroup$

How to evaluate $\displaystyle \lim_{x\to0}\left(\frac{4^x-1}{8^x-1}\right)$ without L'Hopital rule?

When I evaluated this limit I got an indetermination, $\frac{0}{0}$. I learned that in a rational function when one get $\frac{0}{0}$ indeterminated form, one should look for the common terms between numerator and denominator by factoring. But I can't figure out how to find the common terms in this case. Can you help me? Thanks.

$\endgroup$
15
$\begingroup$

Note that $$4^x-1=(2^2)^x-1=(2^x)^2-1=\color{red}{(2^x-1)}(2^x+1)$$ $$8^x-1=(2^3)^x-1=(2^x)^3-1=\color{red}{(2^x-1)}(2^{2x}+2^x+1)$$

$\endgroup$
7
$\begingroup$

Hint We can factor $a^{k x} - 1$ as: $$a^{k x} - 1 = (a^x - 1)(a^{(k - 1) x} + a^{(k - 2) x} + \cdots + a^x + 1).$$

$\endgroup$
1
$\begingroup$

HINT:

As $\lim_{h\to0}\dfrac{e^h-1}h=1,$ and as $ a=e^{\ln(a)}$

$$\lim_{h\to0}\dfrac{a^h-1}h=\ln(a)\cdot\lim_{h\to0}\dfrac{e^{h\ln(a)}-1}{h\ln a}=\ln a$$

Now we know $\ln(b^m)=m\ln(b)$ when both logarithm remain defined

For real calculus we need $a>0$

$\endgroup$
1
$\begingroup$

$$\displaystyle \lim_{x\to0}\frac{4^x-1}{8^x-1}=\lim_{x\to0} (\frac{4^x-1}{x}.\frac{x}{8^x-1})$$

now :

$$\lim_{x\to0}\frac{4^x-1}{x}=\ln 4$$

$$\lim_{x\to0}\frac{x}{8^x-1}=\lim_{x\to0}\frac{1}{\frac{8^x-1}{x}}=\frac{1}{\ln8}$$

so :

$$\lim_{x\to0}\frac{4^x-1}{8^x-1}=\frac{\ln 4}{\ln8}=\frac{2\ln 2}{3\ln 2}=\frac{2}{3}$$

$\endgroup$
  • 1
    $\begingroup$ change 3ln in denominator to 3 ln 2 $\endgroup$ – user12345 Apr 29 '17 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.