Hi blckfin88!
Here are the answers to your questions.
Find the following limits:
(a) lim(a>4) (sqrt(a)-sqrt(4))/(a-4)
Here we can multiply both the numerator and the denominator by
(sqrt(a)+sqrt(4))
This gives
[(sqrt(a)-sqrt(4))(sqrt(a)+sqrt(4))]/(a-4)
But then, in the numerator, we can use the property that
(a+b)(a-b)=a^2+b^2, thus obtaining,
[sqrt(a)^2-sqrt(4)^2)]/(a-4) = (a-4)/(a-4) = 1
Thus the limit here is 1.
(b) lim(x>0) (sin(3x)tan(x))/x^2
Here we use the fact that tan(x)=sin(x)/cos(x). We obtain
[sin(3x)*sin(x)]/(x^2*cos(x))
= [sin(3x)*sin(x)]/(x*x*cos(x))
= (sin(3x)/3x)*(sin(x)/x)*(3/cos(x)) (here I multiplied the num. and
dem. by 3)
Now, we can use the multiplication property of the limits (found in
the following link) in order to find the limit of each term separately
and then multiply them.
Properties for limits of functions
http://www.shu.edu/projects/reals/cont/proofs/limprop.html
Thus we have to find the limits of the following expressions when x
goes to 0:
sin(3x)/3x (1)
sin(x)/x (2)
3/cos(x) (3)
Since in (1) and (2) both denominator and the numerator go to 0 as x
goes to 0, we can use L'Hôpital's rule to find that the limit of both
(1) and (2) as x goes to 0, is 1.
L'Hôpital's Rule
http://www.math.hmc.edu/calculus/tutorials/lhopital/
Since cos(0)=1, then the limit of 3/cos(x) when x goes to 0 is
obviously 3. Finally, the limit of the original expression is simply
then 1*1*3=3.
Determine the following one-sided limits:
(a) lim(x>1-) (x-1)/(abs(x-1))
Since x goes to 1 "from the left" then x is strictly less than 1. Then
(x-1) is strictly less than 0. Therefore, when x is strictly less than
1,
abs(x-1) = -(x-1)
Thus the expression becomes
(x-1)/[-(x-1)] = -1
So -1 is the answer.
(b) lim(x>2+) (x-1)/(x^2-3x+2)
The denominator is a second degree polynomial (a quadratic function).
We can then rewrite it if we know its roots. The formula to calculate
the roots of a quadratic functions is given in the following link:
Quadratic Polynomials
http://www.sosmath.com/algebra/factor/fac08/fac08.html
In this way we find that the roots of the function defined in the
denominator are 1 and 2. We can rewrite the polynomial:
x^2-3x+2 = (x-1)(x-2)
So the original expression becomes
(x-1)/[(x-1)(x-2)]
=1/(x-2)
Since x goes to 2 "from the right, then (x-2) is strictly greater than
0, and therefore the limit of the expression is +Infinity.
Let f (x )= x^2 - 2x - 3.
(a) Find the slope of the secant line between (1, f (1)) and (2, f
(2)).
The definition for secant and tangent lines is given in the following
link:
Secant Line
http://mathworld.wolfram.com/SecantLine.html
In order to calculate this we must first compute f(1) and f(2). We get
that:
f(1) = -4
f(2) = -3
So we have to find the slope of the line connecting the points (1,-4)
and (2,-3). The formula to obtain the slope of a line that connects
two points is given in the following link:
Slope and Graphing Lessons
http://www.purplemath.com/modules/slopgrph.htm
So we calculate the slope as (-3 - (-4))/(2-1) = 1
(b) Find the slope of the tangent line at (1, f (1)) using the
limit definition of the derivative.
The limit definition of the derivative is given in the following link.
I'll write here 'h' instead of 'delta x'.
Derivatives using the limit definition
http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/defderdirectory/DefDer.html
So the derivative of the function
x^2-2x-3
is calculated as
lim (x+h)^2 - 2(x+h) - 3 - x^2 + 2x + 3 / h
h->0
=lim x^2 + 2xh + h^2 - 2x - 2h - 3 - x^2 + 2x + 3 / h
h->0
=lim x^2 + 2xh + h^2 - 2x - 2h - 3 - x^2 + 2x + 3 / h
h->0
=lim 2xh + h^2 - 2h / h
h->0
=lim 2x + h - 2
h->0
=2x - 2
So, at x=1, the derivative is 2*1-2 = 0. So the slope of the tangent
line at (1,f(1)) is 0.
(c) Find the equation of the tangent line (1, f (1)).
Since the slope of the tangent line at (1,f(1)) is 0, then the
equation for the tangent line at this point is simply f(1), which
gives -4.
Recall that the equation for the tanbgent line at point (a,f(a)) is:
t(x) = f'(a)(x-a) + f(a)
Since f'(1) = 0, then the equation of the the tangent line at (1,f(1))
is
t(x) = -4
In the following link, you can find an alternative way to calculate
the tangent line at a point.
Tangent Line
http://www.swt.edu/slac/math/TANGENTLINE/TANGENTLINE.HTML
Find a linear approximation to f (x) = sqrt(x) at x = 64
A linear approximation to a curve at a point is simply the tangent
line at that point. In order to calculate the tangent line, we can use
the equation I gave above. But we must first know the derivative of
the function sqrt(x). The derivative of sqrt(x) is 1/(2sqrt(x)). So
the derivative of sqrt(x) at x=64 is 1/(2*8)=1/16. Then the equation
for the tangent line (or linear approximation) at x=64 is:
t(x) = (1/16)(x-64) + f(64) = (1/16)(x-64) + 8
Use the previous linear approximation and estimate sqrt(65).
Here we have to replace 65 in the equation given above. So
t(65) = (1/16)(65-64) + 8
= 8.0625
Google search strategy used:
properties of limits
://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=properties+of+limits
l'hopital rule
://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=l%27hopital+rule&spell=1
find roots polynomial formula second
://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=find+roots+polynomial+formula+second
secant line
://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=secant+line
limit definition derivative
://www.google.com/search?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=limit+definition+derivative
tangent line
://www.google.com/search?sourceid=navclient&ie=UTF-8&oe=UTF-8&q=tangent+line+equation
I hope this helps! If you have any doubt regading my answer, please
request a clarification before rating it. Otherwise I await your
rating and final comments.
Best wishes!
elmarto |
