Evaluating Limits Analytically: lim[x->0] tan^2(x)/x, etc

[Amadeus]

I really need help! I don't understand how to solve any of these. My teacher wasn't a great help either.

Determine the limit of the trigonometric function (if it exists).

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

2. lim cos x tan x / x
(x -> 0)

For both of those problems, I tired multiplying both top and bottom by X - only to find myself stuck again. Someone, please help and explain it to me!

 

[arthur ohlsten]

1)
tan 2x / x lim x-->0
sin2x/[xcos2x] lim x-->0 =0/0 undefined use L'hopitals rule
d/dx [sin2x] / d/dx[x cos2x] lim x-->0
2cos2x/[ -2xsin2x- cos2x] lim x-->0 = 2/[0-1] =-2 answer



2)
cosxtanx/x lim x-->0
but tan x= sinx/cosx
cosxsinx/[xcosx] lim x-->0
sinx/x lim x--> = 0/0 undefined use L'hospiyals rule
d/dx [sinx] / d/dx x lim x-->0
cosx/1 lim x-->0 = 1 answer


please check math
Arthur

 

[galactus]

For the first one:

\(\displaystyle \L\\\lim_{x\to\0}\frac{tan^{2}x}{x}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{sinx}{x}\cdot\frac{sinx}{1}\cdot\frac{1}{cos^{2}}\)

\(\displaystyle \L\\(1)(0)(1)=0\)

 

[arthur ohlsten]

lim tan^2x / x lim x-->0 but tan x = sin x /cosx
lim sin^x sin x /[x cos^2x] as x-->0

sin 0=0
cos 0=1

limit becomes 0*0/[0*1*1]
becomes 0/0 undefined

by L'Hopitals rule
d/dx sin^2x/ d/dx [x cos^2x]=2 sin x cosx /-[x 2 cosxsin x +cos^2x ]
lim x-->0
2*0*1/[-0*2*1*0+1]
0/[0+1]
0 answer
Arthur

 

[soroban]

Re: Evaluating Limits Analytically: lim[x->0] tan^2(x)/x,

Hello, Amadeus!

From the nature of these problems, I don't think L'Hopital's Rule is appropriate.

You're expected to know this theorem: \(\displaystyle \L\:\lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1\)

. . and this basic identity: \(\displaystyle \L\:\tan\theta \:=\:\frac{\sin\theta}{\cos\theta}\)

\(\displaystyle \L 1)\;\;\lim_{x\to0}\frac{\tan^2x}{x}\)

We have: \(\displaystyle \L\:\frac{\left(\frac{\sin^2x}{\cos^2x}\right)}{x} \;=\;\sin x\,\cdot\,\frac{\sin x}{x}\,\cdot\,\frac{1}{\cos^2x}\)

Then: \(\displaystyle \L\:\lim_{x\to0}\left[\sin x\,\cdot\,\frac{\sin x}{x}\,\cdot\,\frac{1}{\cos^2x}\right]\)

. . \(\displaystyle \L= \;\underbrace{\left[\lim_{x\to0}\sin x\right]}\,\underbrace{\left[\lim_{x\to0}\frac{\sin x}{x}\right]}\,\underbrace{\left[\lim_{x\to0}\frac{1}{\cos^2x}\right]}\)
. . . . . . . . . .\(\displaystyle \downarrow\) . . . . . . . . . .\(\displaystyle \downarrow\) . . . . . . . . . .\(\displaystyle \downarrow\)
. . \(\displaystyle \L=\;\;\;\;\;\;0\;\;\;\;\;\cdot\;\;\;\;\;1\;\;\;\;\;\cdot\;\;\;\;\;\frac{1}{1^2}\)

. . \(\displaystyle \L=\;\;\;0\)

\(\displaystyle \L 2)\;\;\lim_{x\to0}\frac{\cos x\cdot\tan x}{x}\)

We have: \(\displaystyle \L\:\lim_{x\to0}\frac{\cos x\,\cdot\,\frac{sin x}{\cos x}}{x} \;=\;\lim_{x\to0}\frac{\sin x}{x} \;=\;1\)