### Question:

### Evaluate the limits:

lim(theta symbol–>0-) csc theta symbol

lim(theta–>pi/2) 1/2 tan theta

Calculate this limit at infinity (+/- infinity) for the following function:

f(x) = e^x+e^-x/2

### Answer:

### When you say f(x), calculus and programmable calculators are saying “y”. When they ask you for a limit, it is a trick question meaning find y. They then give you an x value and say as x approaches 5 (for instance) what is the limit? You then would plug 5 in for x ( which my programs do) and the answer is the limit is the value of “y”. Most of the time they trick you again do to the function they give you which is always a fraction such as (3x+7)/(x^2-4), (always a division sign). Then they ask you as x–>2 (as x approaches 2) what’s the limit. You’ll notice that when you plug in 2 for x you get a zero in the denominator which becomes an “undefined” answer.

### If you then (like my program does) add some very small number to 2 such as .001 to get 2 + .001 = 2.001 or take a small number away from 2 to get 2-.001 = 1.999. Setting that equal to x then you actually get a number and that is the limit. I would actually graph the original function quickly on the calculator so I could see the picture of the function and then use the cursor to scroll back and forth at the x value they give you to do things like what happens when x approaches this from the right or left, to give me any clue of what their talking about.

### Same thing with infinity. Graph the function first and then logically think about what they are asking. Also start to make things perfectly clear in your calculation examples so there is no question of what you’re asking.

### For instance, you have:

### f(x) = e^x+e^-x/2 which to a calculator, and me, means

### f(x) = e^(x)+e^(-x)/2

### and when you mean to say:

### f(x) = e^(x)+e^(-x/2)

### The problem says calculate the limit at infinity. I would graph the function on the calculator. You can see it’s just a parabola (valley looking, because no minus sign before it).

### Here’s what they might be looking for:

### f(x) = e^(x)+e^(-x/2)

### Convert to:

### = x*ln(e)+(-x/2)*ln(e)

### factor out the ln(e)

### = ln(e)[x+(-x/2)]

### ln(e) = 1 So:

### (1)[x+(-x/2)]

### The limit as x-> ∞

### = [ ∞ + (-∞/2)]

### = ∞

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