hello every body

i need ahelp

http://www.mathmontada.net/vb/upload...1369571893.gif

Printable View

- May 26th 2013, 08:32 AMabualabedlimit
hello every body

i need ahelp

http://www.mathmontada.net/vb/upload...1369571893.gif - May 26th 2013, 09:01 AMMINOANMANRe: limit
the limit = 2

try to find it.....show us some of your work... - May 26th 2013, 11:27 AMabualabedRe: limit
It is easy to get the final answer, but the procedure remain....

- May 26th 2013, 12:08 PMShakarriRe: limit
Try L'hospitals rule

- May 26th 2013, 02:57 PMx3bnmRe: limit
We've to find the limit of:

$\displaystyle \lim_{x \to 0}\frac{\sin(\tan(2x))}{\log(x+1)}$

It's in indeterminate form means the function is in the form of $\displaystyle \frac{0}{0}$.

So using I'Hopital's rule we get:

$\displaystyle \begin{align*}& \lim_{x \to 0}\frac{\sin{(\tan(2x))}}{\log(x+1)}\\=& \lim_{x \to 0} \frac{2\cos{(\tan(2x))}\sec^{2}(2x)}{\frac{1}{x+1} }\\=& \lim_{x \to 0} 2\cos{(\tan(2x))}\sec^{2}(2x) (x+1)\\=& (2)(1)(1)(0+1)\\=& 2......\text{[Answer]}\end{align*}$

You can check the answer here:

\lim_{x \to 0}\frac{\sin(\tan(2x))}{\log(x+1)} - Wolfram|Alpha