Trig Limit Question: Solving for Undefined Result | Help Needed

[helpcalc]

Could someone help me with this one. My teacher says I'm wrong when I solved this limit as undefined. thanks!

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

 

[Bohrok]

If you mean (sin(x)tan(x)-1)/(4x), then it would be undefined. I have a feeling that the -1 is causing a problem since you may be missing parentheses that would change the problem and let a limit exist.

 

[n!kofeyn]

If the limit you are asking about is
\lim_{x\to 0} \frac{\sin x \tan x - 1}{4x}
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.

 

[helpcalc]

thanks for the responses but when i graph this function at 0 the function is zero. It looks like it is continuous and goes to zero from the left and right but i can't find a mathematical way to solve it.

 

[Cyosis]

Well you must have drawn the graph incorrectly. For example what value did you get for x=0.1 and x=-0.1?

 

[helpcalc]

for -.01 the limit is .0025 and for .01 the limit is -.0025. looking at a graphing calculator using the table function, it looks like it converges to 0.

 

[Cyosis]

I see, that means we're talking about different functions and so are the other posters in this thread.

The limits you describe correspond with the function:

<br /> x \left(\frac{\sin x \tan x -1}{4}\right)<br />

We assumed you meant:

<br /> \frac{\sin x \tan x - 1}{4x}<br />

Could you clarify which one it is?

 

[helpcalc]

my mistake on the calculator. it is definitely the second one. looking at the graph my teacher must be wrong (I hope!) thanks for your help!

 

[Cyosis]

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