Calc Midterm 2 Review

Question 1

How do you find higher order derivatives?

Answer 1

You simply take the derivative of the derivative.

examples like f’(x) to f’‘(x)

Question 2

Explain the solution to:

Limit as x->0 of sin(ax) / x

Answer 2

So there’s a constant given in the sin operator. We know that in order for the whole process to become one we need to multiply the function by the constant/constant: a/a

(sin(ax) / x) · a/a -> asin(ax) / ax

Now we need to remove the a in the numerator which only leaves us with sin(ax) / ax. From this we can identify a similarity with (lim. x → 0 of sin(x) / x = 1)

After multiplying the both sides by a you would get:

a: lim. x-> 0 of sin(ax) / ax
or
a: 1=a

Question 3

What are the two trig. limit identities?

Answer 3

lim. x → 0 of sinx/x = 1
lim. x → 0 of 1-cosx / x = 0

Question 4

Explain the logic behind:

limit x → 0 of sinx / x

Answer 4

We can utilize the Squeeze Theorem which takes our unit circle triangles of:
tanx/2 ≧ x/2 ≧ sinx/2

You would then multiply each side by 2/sinx and flip the reciprocal of each value. This will give us

cosx ≧ sinx/x ≧ 1

@x = 0, cos(0) = 1 gives us

1﹥sinx/x ﹥1 which makes our limit:

limit x → 0 of sinx/x = 1

Question 5

Explain the logic behind:

limit x → 0 of 1 - cosx / x

Answer 5

Understanding that Lim. x → 0 of sinx/x = 1, we start by trying to get the x out of the denominator by multiplying our function by the conjugate of 1 - cosx:

1 - cosx/x · (1 + cosx)/(1 + cosx) → 1 - cos²x into sin²x then split the function into two digestible parts:

sin²x / x(1 + cosx) → sinx/x · sinx/(1+cosx)

Applying the limit operator, we can now find the limit of both side of the function:

lim. x → 0 of sinx/x = 1
lim. x → 0 of 1-cosx / x = 0

0 = 0

Question 6

Explain the solution to:

limit x → 0 of sinax / sinbx

Answer 6

Start by using the procedure for limit x → 0 sinax / x, where we learned this is a