I want to check my answers

Here's the last one

lim x→ 0   [3x - tan^2(7x)] / (2x)      you have the first part correct........let's look at this part

lim x→ 0   [ - tan^2(7x)] / (2x)     note we can write this as.....

lim x→ 0    -[ tan(7x)] * [tan(7x)] / (2x)...note, we can use a "trick" here for the second expression.....multiply the top and bottom by 7....this gives

lim x→ 0    -[ tan(7x)] * [7tan(7x)] / (7*2x)  =  lim x→ 0   -[ tan(7x)] * [7tan(7x)] / (2 * 7x) =

lim x→ 0   (-7/2)[ tan(7x)] * [tan(7x)] / ( 7x)      and concentrating on  [tan(7x)] / ( 7x), we can write

sin(7x)/[cos(7x)*7x] = sin(7x)/7x *1/ cos(7x)]   and using an identity,  lim x → 0   sin(7x)/7x) = 1, and

lim x→ 0   1/ cos(7x)  = 1...so we're left with

lim x→ 0   (-7/2)[ tan(7x)]   and as x → 0, tan(7x) → 0    so.... (-7/2)* 0 = 0

And your final answer of 3/2 is correct.....!!!!

4k² - 8 = 4k - 8

k² = k

k = 0 or k = 1

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Couldn't read your third one, but the first two are:

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Here's another way to do (a)

lim x → 0⁺  √[1 - cos(x)] / x        multiply top and bottom by √[1 + cos(x)]   so we have

lim x → 0⁺  √[1 - cos^2(x)] /[ (x) √[1 + cos(x)]    =

lim x → 0⁺  sin(x) /[ (x) √[(1 + cos(x)]  =

lim x → 0⁺  [sin(x) / (x)] * [1/ √[1 + cos(x)]      and, using an identity...... lim x → 0⁺  [sin(x) / (x)]  = 1

So we have

lim x → 0⁺  [1/ √[1 + cos(x)]       and taking the limit, we have

[1/ √[1 + 1]  =  1 / √2

Alan I don't understand  " b " 

this the one you cant read 

Here's (b) again

lim x → 3     (x - 3) / (1/x - 1/3)^2

Note that, getting a common denominator,   (1/x - 1/3)   becomes (3 - x) / 3x......so we have

lim x → 3     (x - 3) / [(3 - x) / 3x]^2        and notice that we can write the numerator as -(3 - x)......and this "cancels"  with one of the linear factors of (x - 3)^2 in the denominator leaving just -1 in the numerator....so we have...

lim x → 3     -1 ÷ [(3 - x) / [3x]^2].......so, if we had   -1 ÷ (5/6)    we would just invert the second fraction and change to multiplication.....doing the same thing here, we have.....

lim x → 3    -1 x   [3x]^2 / (3 - x)  =

lim x → 3    - [3x]^2 /(3 - x)

Note, that when x → 3^-, this function is negative. And when x  → 3⁺ , it's positive. Thus, the limit does not exist.

What is the number on "x" in the "tan^2"  term??

tan^2(7x)  ... 

chris .. 

the question is  lim when x approaches to 3   not zero 

Oops....I'll do an edit to the original post.....sorry!!!!

so my answer is correct ! 

Thank you CPhill

^^^^^^ ? ? 

The first two seem ok (as far as I can see, though I can't see the final expression exactly!).

For the third one:

$$y=5\sqrt{x}-x^{-3/2}+\pi^{3/4}$$

$$\frac{dy}{dx}=\frac{5}{2\sqrt{x}}+\frac{3}{2x^{5/2}}$$

$$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}(5+\frac{3}{x^2})$$

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Thank you Alan