Limits using Trig Functions

• Sep 22nd 2008, 04:17 AM
Roxanne
Limits using Trig Functions
Can someone help me with this: Find the limit if the limit exists:

lim x-tanx
x->0 sinx

lim 1 - sine^2 (1)^2
x->-2- t (t)
• Sep 22nd 2008, 05:08 AM
mr fantastic
Quote:

Originally Posted by Roxanne
Can someone help me with this: Find the limit if the limit exists:

lim x-tanx
x->0 sinx

Mr F says: $\displaystyle {\color{red} \lim_{x \rightarrow 0} \frac{x - \tan x}{\sin x} = \lim_{x \rightarrow 0} \left( \frac{x}{\sin x} - \frac{1}{\cos x} \right) = \lim_{x \rightarrow 0} \frac{x}{\sin x} - \lim_{x \rightarrow 0} \frac{1}{\cos x}}$

lim 1 - sine^2 (1)^2 Mr F says: This is incomprehensible.
x->-2- t (t)

..
• Sep 22nd 2008, 07:02 AM
Roxanne
Limits using trig functions
find limit if limit exist:

lim 1/t^2 sin^2 (t/2)
x->0
• Sep 22nd 2008, 09:39 AM
Soroban
Hello, Roxanne!

If no one is answering, there is confusion in your statement . . .

I must assume it says: .$\displaystyle \lim_{x\to0}\frac{\sin^2\!\frac{x}{2}}{x^2}$

In the denominator, multiply by $\displaystyle \frac{4}{4}$

. . $\displaystyle \lim_{x\to0}\,\frac{\sin^2\!\frac{x}{2}}{\frac{4}{ 4}\cdot x^2} \;=\;\lim_{x\to0}\:\frac{\sin^2\!\frac{x}{2}}{4\cd ot\frac{x^2}{4}} \;=\;\lim_{x\to0}\:\frac{1}{4}\cdot\frac{\sin^2\!\ frac{x}{2}}{(\frac{x}{2})^2}$ .$\displaystyle = \;\frac{1}{4}\cdot\lim_{x\to0} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2 \;=\;\frac{1}{4}\cdot1^2 \;=\;\frac{1}{4}$

• Sep 22nd 2008, 01:12 PM
mr fantastic
Quote:

Originally Posted by Roxanne
find limit if limit exist:

lim 1/t^2 sin^2 (t/2)
x->0

Do you mean $\displaystyle \frac{1}{t^2} \, \sin^2 \frac{t}{2}$ ? If so note that:

$\displaystyle \frac{1}{t^2} \, \sin^2 \frac{t}{2} = \frac{\sin^2 \frac{t}{2}}{t^2} = \frac{1}{4} \, \left( \frac{\sin \frac{t}{2}}{\frac{t}{2}} \right) \, \left( \frac{\sin \frac{t}{2}}{\frac{t}{2}} \right)$.