I'm struggling with the following problem: Lim as X-->0 of (tan^2 X)/X Thanks
K Ken New member Joined Sep 7, 2007 Messages 9 Sep 12, 2007 #1 I'm struggling with the following problem: Lim as X-->0 of (tan^2 X)/X Thanks
pka Elite Member Joined Jan 29, 2005 Messages 11,976 Sep 12, 2007 #2 Take note that: \(\displaystyle \L \frac{{\tan ^2 (x)}}{x} = \left( {\frac{{\sin (x)}}{x}} \right)\left( {\frac{1}{{\cos ^2 (x)}}} \right)\left( {\sin (x)} \right).\)
Take note that: \(\displaystyle \L \frac{{\tan ^2 (x)}}{x} = \left( {\frac{{\sin (x)}}{x}} \right)\left( {\frac{1}{{\cos ^2 (x)}}} \right)\left( {\sin (x)} \right).\)
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 12, 2007 #3 You can break it up into fractions: \(\displaystyle \L\\\lim_{x\to\0}\frac{tan^{2}(x)}{x}=\lim_{x\to\0}\frac{sin(x)}{x}\cdot\lim_{x\to\0}\frac{sin(x)}{1}\cdot\lim_{x\to\0}\frac{1}{cos^{2}(x)}\) You should see a familiar limit in sin(x)/x.
You can break it up into fractions: \(\displaystyle \L\\\lim_{x\to\0}\frac{tan^{2}(x)}{x}=\lim_{x\to\0}\frac{sin(x)}{x}\cdot\lim_{x\to\0}\frac{sin(x)}{1}\cdot\lim_{x\to\0}\frac{1}{cos^{2}(x)}\) You should see a familiar limit in sin(x)/x.
K Ken New member Joined Sep 7, 2007 Messages 9 Sep 12, 2007 #4 A limit problem Thank you for your help. I need to digest your responses a bit.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Sep 13, 2007 #5 You just need to remember the classic identity \(\displaystyle \L\\\frac{sin(x)}{cos(x)}=tan(x)\). That's all it is.
You just need to remember the classic identity \(\displaystyle \L\\\frac{sin(x)}{cos(x)}=tan(x)\). That's all it is.