Solve the limit without using L'Hôpital's rule

Question

$\displaystyle\lim\limits_{x \to 0} \frac{8(\frac 12 + x)^8-8(\frac 12)^8}{x}$

My attempt at the problem: I tried using substitution, $h = \frac 12 +x$, but I ended up where I began. Any suggestions would be great.

Answer 1

HINT

Try to divide both, numerator and denominator by $x$. Keep in mind that $1/x$ goes to infinity as $x$ goes to zero.

(I really like Anurag's answer, using the definition of the derivative is a really good method for this particular limit)

Answer 2

Hint

Recall $$f'(a)=\lim_{x \to 0} \frac{f(a+x)-f(a)}{x}.$$

So this limit is the derivative of $f(t)=8t^8$ at $t=1/2$.

Answer 3

Use the binomial theorem on the first term. The numerical term cancels, and the only term not containing an $x$ is $8\times8(\frac 12)^7,$which must be the limit as $x\rightarrow0$

Answer 4

If you are not allowed to use derivatives, then use algebra: \begin{align} a^8-b^8 &=(a^4-b^4)(a^4+b^4)\\ &=(a^2-b^2)(a^2+b^2)(a^4+b^4)\\ &=(a-b)(a+b)(a^2+b^2)(a^4+b^4) \end{align} In your case $a=\frac{1}{2}+x$ and $b=\frac{1}{2}$, so you get \begin{align} \lim_{x \to 0} \frac{8(\frac 12 + x)^8-8(\frac 12)^8}{x} &=\lim_{x\to0} 8\frac{ x \Bigl(\frac{1}{2}+x+\frac{1}{2}\Bigr) \Bigl((\frac{1}{2}+x)^2+\frac{1}{4}\Bigr) \Bigl((\frac{1}{2}+x)^4+\frac{1}{16}\Bigr) }{x}\\[6px] &=\lim_{x\to0} 8(x+1) \Bigl((\tfrac{1}{2}+x)^2+\tfrac{1}{4}\Bigr) \Bigl((\tfrac{1}{2}+x)^4+\tfrac{1}{16}\Bigr) \end{align} Now you can plug in $x=0$.