evaluate lim [tan(x)-sin(x)] / sin^2(x) as x->0

[mathy]

I want to evaluate:

. . .1) lim [x->0] [tan(x)-sin(x)] / sin^2(x)

What do I do? OK, so we know that the lim as x-->1 of sin(x)/x=1..........???

The other limit I didn't know how to evaluate was:

. . .2) lim [x->0] (x^2)/[(sin^2)(2x)]

Can anyone help?

 

[arthur ohlsten]

1) two ways to do this

x-->0 [tanx-sinx]/sin^2x = 0/0 undefined
take derivative of numerator and denominator

x-->0 [(1/cos^2x) - cosx] /2 sinx cosx = [1-1]/0
0/0 undefined take derivative as before

x-->0 [cos^-2 x -cosx]/2 sinx cosx
x-->0 cosx[cosx -1] / sin 2x
x-->0 [-cosxsinx-[cosx-1]sinx] /2cos 2x = [0-0]/ 2
answer 0

second way

[ tanx -sinx]/ sin^2x = [sinx - sinx cosx]/cosx / sin^2x
[tanx-sinx]/sin^2x= sinx[1-cosx]/sin^2x
[tanx-sinx]/sin^2x=[1-cosx]/sin x lim x-->0
= 0/0 undefined take derivative of numerator and denominator
x-->0 sinx/cosx = 0 answer

=================================================
2
lim x-->0 x^2/sin^2 2x =0/0 undefined
take derivative of numerator and denominator L'Hopitals rule
x-->0 2x/[ 4 sin 2x cos 2x ] = 0/0 undefined l'hopitals rule

x-->0 1/[2 (2 sin^2 2x+2 cos^2 2x ] = 1/[4]
1/4 answer

Arthur
please check math the technique is right

 

[galactus]

#2:

You can use L'Hopital's rule:

Differentiate numerator and denominator twice:

\(\displaystyle \L\\\frac{d}{dx}[x^{2}]=2x; \;\ \frac{d}{dx}[2x]=2\)

\(\displaystyle \L\\\frac{d}{dx}[sin^{2}(2x)]=4sin(2x)cos(2x); \;\ \frac{d}{dx}[4sin(2x)cos(2x)]=16cos^{2}(2x)-8\)

\(\displaystyle \L\\\frac{\lim_{x\to\0}2}{\lim_{x\to\0}(16cos^{2}(2x)-8)}=\frac{2}{8}=\frac{1}{4}\)

 

[mathy]

thanks for your help!!... i see how you got the answers
but i was also wondering how we could find them without using derivatives, and just the limit laws

 

[skeeter]

\(\displaystyle \L \frac{\tan{x}-\sin{x}}{\sin^2{x}} = \frac{1-\cos{x}}{\sin{x}\cos{x}}\)

multiply numerator and denominator by 1 + cosx ...

\(\displaystyle \L \frac{1-\cos^2{x}}{\sin{x}\cos{x}(1+\cos{x})} = \frac{\sin^2{x}}{\sin{x}\cos{x}(1+\cos{x})} = \frac{\sin{x}}{\cos{x}(1+\cos{x})}\)

now take the limit as x -> 0

 

[galactus]

Here's #2 without derivatives.

\(\displaystyle \L\\\frac{x^{2}}{sin^{2}(2x)}=\left(\frac{x}{sin(2x)}\right)^{2}\)...[1]

You know about the limit of \(\displaystyle \frac{x}{sin(x)}\). Find the limit of \(\displaystyle \frac{x}{sin(2x)}\)

\(\displaystyle \frac{1}{2}\lim_{x\to\0}\frac{2x}{sin(2x)}\)

Let t=2x.

\(\displaystyle \L\\\lim_{t\to\0}\frac{t}{sin(t)}=1\)

(1/2)(1)=1/2

Sub back up into [1]

\(\displaystyle \H\\(\frac{1}{2})^{2}=\frac{1}{4}\)

L'Hopital is usually an easy way if one has an indeterminate form. I think it is better to see if you can do it, first, without L'Hopital. Although, I have been guilty of using it outright.

 

[mathy]

THANKS!