1. ## limits

$lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

$lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

$lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

$lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

$lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

can i verify the solution..thanks

2. [quote=ashes;127090] $lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

$lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

$lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

$lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

$lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

can i verify the solution..thanks...for the first one because of l'hopitals we have $\lim_{x \to 0}\frac{xcos(x)-x}{sin(x)-x}=\lim_{x \to 0}\frac{-xsin(x)+cos(x)-1}{cos(x)-1}$...applying L'hopitals rule again we get $\lim_{x \to 0}\frac{-xcos(x)-sin(x)-sin(x)}{-sin(x)}$ splitting we get $\lim_{x \to 0}\frac{-xcos(x)}{-sin(x)}-2$...and finally differntiatting on the first one we et $\lim_{x \to 0}\frac{xsin(x)-cos(x)}{-cos(x)}-2=-1$

3. [quote=Mathstud28;127109]
Originally Posted by ashes
$lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

$lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

$lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

$lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

$lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

can i verify the solution..thanks...for the first one because of l'hopitals we have $\lim_{x \to 0}\frac{xcos(x)-x}{sin(x)-x}=\lim_{x \to 0}\frac{-xsin(x)+cos(x)-1}{cos(x)-1}$...applying L'hopitals rule again we get $\lim_{x \to 0}\frac{-xcos(x)-sin(x)-sin(x)}{-sin(x)}$ splitting we get $\lim_{x \to 0}\frac{-xcos(x)}{-sin(x)}-2$...and finally differntiatting on the first one we et $\lim_{x \to 0}\frac{xsin(x)-cos(x)}{-cos(x)}-2=-1$

The second and third one are just more simple applications of L'hopitals if I have time tonight I will show them...here is the fourth one $\lim_{r \to 1}\frac{ar^{n}-a}{r-1}$....since it is $\frac{0}{0}$...we can apply L'hopital's rule to get $\lim_{r \to 1}\frac{anr^{n-1}}{1}$=an

4. ## And I will give you a hint for the last one

Originally Posted by ashes
$lim_{x\rightarrow0}\ \frac{x(cosx-1)}{sinx-x}$

$lim_{x\rightarrow0}\ \frac{\sqrt{a(a+x)}-a}{x}\ , a>0$

$lim_{x\rightarrow{\pi/2}}\ \frac{ln(sinx)}{(\pi-2x)^2}$

$lim_{r\rightarrow1}\ \frac{a(r^n-1)}{r-1} , n\ is\ a\ positive\ integer$

$lim_{x\rightarrow0}\ ( \frac{1}{sinx}-\frac{1}{x})$

can i verify the solution..thanks
if you combine them you have $\lim_{x \to 0}\frac{x-sin(x)}{xsin(x)}$....hmm....I wonder what you could use on that...*cough*L'hopital's*cough*