- #1

helpcalc

- 4

- 0

lim as x goes to zero of (sin x tan x -1)/4x

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- Thread starter helpcalc
- Start date

- #1

helpcalc

- 4

- 0

lim as x goes to zero of (sin x tan x -1)/4x

- #2

Bohrok

- 867

- 0

- #3

n!kofeyn

- 537

- 3

[tex]\lim_{x\to 0} \frac{\sin x \tan x - 1}{4x}[/tex]

then the limit is undefined, and your teacher is wrong. The way to determine this is to look at the limit as x approaches 0 from the left and then from the right.

The limit of the numerator as x approaches 0 is -1. As x approaches 0 from the left, x is negative and so the limit will be positive infinity. As x approaches 0 from the right, x is positive, so you will get negative infinity. Therefore, the limit does not exist since the left and right handed limits aren't equal.

- #4

helpcalc

- 4

- 0

- #5

Cyosis

Homework Helper

- 1,495

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- #6

helpcalc

- 4

- 0

- #7

Cyosis

Homework Helper

- 1,495

- 0

The limits you describe correspond with the function:

[tex]

x \left(\frac{\sin x \tan x -1}{4}\right)

[/tex]

We assumed you meant:

[tex]

\frac{\sin x \tan x - 1}{4x}

[/tex]

Could you clarify which one it is?

- #8

helpcalc

- 4

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- #9

Cyosis

Homework Helper

- 1,495

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If it's the second one then the limit does not exist.

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