We're reviewing limits from the previous chapter, but I'm still having some difficulty with some problems:

lim tan(pi(x/4)) as x-> -7

lim (x^2 - x^3(cos(1/x))/3x^2 as x-> 0

I know the answer to the first problem is 1 and to the second problem it's 1/3 but I'm not sure how to get these answers. Any help is greatly appreciated! :)

The first step in any limit question is to actually substitute the approach value into the expression.

If you get a real number as an answer, that is it, you are done
so Lim tan(πx/4) as x ---> -7
= lim tan(-7π/4)
= 1

in lim (x^2 - x^3(cos(1/x)/(3x^2) as x ---> 0
= lim (1-xcos(1/x))/3

no matter what cos(1/x) is , xcos(1/x) ---> 0 as x-->0
(multiplying anything by zero, is zero!)

so lim (1-xcos(1/x))/3
= (1 - 0)/3 = 1/3

Reading these answers makes a whole lot more sense after a night's sleep, haha. Thank you so much!

Sure! I'll explain how to find the limits for both of these problems.

Problem 1:

To find the limit of tan(pi(x/4)) as x approaches -7, we can use the fact that the limit of tan(x) as x approaches 0 is 0. Since -7/4 is not equal to 0, we need to make a substitution by multiplying the function by 1 to get a limit format that will cancel out the variable in the argument of the trigonometric function.

Let's substitute y = x + 7. As x approaches -7, y approaches 0.

Therefore, our original limit can be rewritten as:

lim tan(pi((y+7)/4)) as y-> 0

Next, we can simplify the expression inside the tangent function:

lim tan(pi(y/4 + 7/4)) as y-> 0

Using the trigonometric identity tan(a+b) = (tan(a) + tan(b))/(1 - tan(a)tan(b)), we can split the expression inside the tangent function:

lim (tan(pi(y/4)) + tan(pi(7/4))) / (1 - tan(pi(y/4))tan(pi(7/4))) as y-> 0

Now, we can evaluate the individual limits inside the tangent functions:

lim tan(pi(y/4)) as y-> 0 = tan(0) = 0
lim tan(pi(7/4)) as y-> 0 = tan(pi(7/4)) = 1

Using the evaluated limits, our expression becomes:

lim (0 + 1) / (1 - 0 * 1) as y-> 0

This simplifies to:

lim 1 / 1 = 1

Therefore, the limit of tan(pi(x/4)) as x approaches -7 is 1.

Problem 2:

To find the limit of (x^2 - x^3(cos(1/x))/3x^2) as x approaches 0, we can use algebraic simplifications and the properties of limits.

First, let's simplify the expression by factoring out x^2:

lim (x^2(1 - x(cos(1/x)))/3x^2) as x-> 0

Next, we can cancel out the x^2 terms:

lim (1 - x(cos(1/x)))/3 as x-> 0

Now, let's consider the behavior of the expression x(cos(1/x)) as x approaches 0.

Since the cosine function is bounded between -1 and 1, we have:

-1 <= cos(1/x) <= 1
-1 * x <= x(cos(1/x)) <= 1 * x
-x <= x(cos(1/x)) <= x

As x approaches 0, both -x and x approach 0. Therefore, x(cos(1/x)) approaches 0.

Now, we can simplify the expression further:

lim (1 - x(cos(1/x))/3) as x-> 0

Since x(cos(1/x)) approaches 0, our expression becomes:

lim (1 - 0)/3 as x-> 0

This simplifies to:

lim 1/3 = 1/3

Therefore, the limit of (x^2 - x^3(cos(1/x))/3x^2) as x approaches 0 is 1/3.

I hope this explanation helps! Let me know if you have any further questions.