1) lim (5-sqrt[4+3x]) / (7-x), x going toward 7
2) lim (tan9x) / (1-cos7x), x going toward 0
3) lim (5sqrt[x]-2) / (x-32), x going toward 32, and the 5 is part of the radical so i guess its x^(1/5)
Hello,
$\displaystyle \lim_{x \to 7} ~ \frac{5-\sqrt{4+3x}}{7-x}$
Multiply by $\displaystyle \frac{5+\sqrt{4+3x}}{5+\sqrt{4+3x}}$
then use identity $\displaystyle (a-b)(a+b)=a^2-b^2$ to simplify the top =)
$\displaystyle \lim_{x \to 0} ~ \frac{\tan (9x)}{1-\cos(7x)}=\lim_{x \to 0} ~ \frac{\tan(9x)}{x} \cdot \frac{x}{1-\cos(7x)}=\lim_{x \to 0} ~ 9 \cdot \frac{\tan(9x)}{9x} \times \frac 17 \cdot \frac{7x}{1-\cos(7x)}$2) lim (tan9x) / (1-cos7x), x going toward 0
These should be known limits :/
$\displaystyle \lim_{x \to 32} ~ \frac{\sqrt[5]{x}-2}{x-32}$3) lim (5sqrt[x]-2) / (x-32), x going toward 32, and the 5 is part of the radical so i guess its x^(1/5)
You can note that $\displaystyle x-32=(\sqrt[5]{x})^5-2^5$
And in general, $\displaystyle a^k - b^k = (a-b)(a^{k-1}+a^{k-2}b+a^{k-3}b^2 + ... +b^{k-1})$
(here is what I've done about this formula : http://www.mathhelpforum.com/math-he...484-k-b-k.html)