calculus help with limits, differentiating?

Question

i think i understand how to do these, but if someone could show the work for the answers and stuff id greatly appreciate it.

7) find the limit (as x goes to 5) of (x^2-25)\(x-5)

8) find the limit (as h goes to 0) of (5(.5+h)^4-5(.5)^4)\h

9) find the slope of the normal line to y = x + cosxy at (0,1)

10) if f(x) is continuous and differentiable, and: f(x) = ax^4+5x (when X is less than or equal to 2), and f(x) = bx2-3x (when X is greater than 2), then find b.

thanks!

for #8, that backwards slash means DIVIDE. my bad haha, i typed in the wrong one =P

lmao gary, you made me laugh with some of the stuff you wrote. in a good way though. the .5 i wrote is actually (1/2), but i didnt want to write any more parenthesis.

9, yes, im positive. she couldve meant y = x + cos(xy) at the point (0,1) but she didnt write it with xy in parenthesis.

by bx2 i meant bx^2, i left out the power thing, sorry.

Answer 1

For 7), take the derivative of the numerator and denominator separately.

lim as x -> 5 of (2x)/1

the limit is 10.

Check: for x = 5.01, the function evaluates to 10.01.
for x = 4.99, the function evaluates to 9.99.

Easier: factor the numerator into (x + 5)(x - 5) and cancel the x - 5. That leaves you with x + 5 FOR ALL VALUES OF X, not just the limit.

Gopal's solution to 7) is wrong because Gopal cannot add 5 + 5.

I can't read 8), and it contains a backward slash, which is not a mathematical symbol. 8) looks wrong. Why have you expressed a constant as 5(.5)^4 instead of multiplying it out? Anyway, taking the derivative of the numerator and denominator separately gives you

20(.5 + h)^3

As h goes to zero, the limit is 20(.5)^3

Gopal's solution for 8) is wrong because Gopal took the derivative wrong.

Are you sure you wrote 9) correctly?

For 10) to be differentiable at x = 2, the slopes of the two functions must be equal at x = 2, whether you approach from the left or from the right.

for x <= 2, f'(x) = 3ax^3 + 5
At x = 2, f'(2) = 24a + 5

For the second function, what does "bx2" mean?

For the functions to be continuous, they must evaluate to the same value at x = 2. The first function evaluates to 16a + 10 at x = 2. The second function is anyone's guess.

Answer 2

7.lt (x+5)(x-5)/(x-5)
lt x>5 (x+5)=15

8.use L hospitals rule
5*4h^3/1=0 as h >0

9.y'=1-sinxy*(x+y')at (0,1)
y'=1-sin(0)(y)(0+y')
=1
line normal to this will have a slope of -1
equation
y-1=-1(x-0)
x+y=1

10.when x is less than 2
f(x)=ax^4+5x
ax^4+5x<16a+10
f(x)=b(2)^2-6 when x>2
16a+10

Answer 3

For 7

The top should factorise to (x+5)(x-5) and cause the elimination of the x-5

This will leave a equivilant sum of x+5

Then substitute x->5

which gives a limit of 5+5 = 10

Answer 4

observe that 4x - x^2 = x(4 - x) = x(2 + sqrt(x))(2 - sqrt(x)) now sub in to grant: lim (2 - sqrt(x)) / [x(2 + sqrt(x))(2 - sqrt(x)) (2 - sqrt(x)) / (2 - sqrt(x)) cancels to a million as long as x =/= 4, and is a detachable discontinuity you are able to substitute in 4 for x in what's left to locate the shrink a million / [x(2 + sqrt(x)] as x ==> 4 is a million / [4(2 + 2)] = a million/sixteen shrink as x ==> 4 = a million/sixteen once you initially substitue 4 for x, you get 0 / 0, that's indeterminate, telling you which you will seek for components of 0 to cancel..