Find the following limits algebraically or explain why they don’t exist.
lim x->0 sin5x/2x
lim x->0 1-cosx/x
lim x->7 |x-7|/x-7
lim x->7 (/x+2)-3/x-7
lim h->0 (2+h)^3-8/h
lim t->0 1/t - 1/t^2+t
a lot of your questions contain ambiguous typing
e.g. the second last one probably says
lim ( (2+h)^3 - 8)/h as h --> 0
You could expand the top, or recognize it as a difference of cubes
recall A^3 - B^3 = (A-B)(A^2 + AB + B^2)
= lim [(2+h - 2)((2+h)^2 + 2(2+h) + 4) )/h , h--->0
= lim (2+h)^2 + 2(2+h) + 4 , h--> 0
= 4 + 4 + 4 = 12
for the first one
I will again assume you meant
lim sin (5x) / (2x) , as x --> 0
recall that lim sinØ/Ø = 1 as Ø -->0
so let's "construct" this pattern
multiply top and bottom by (5/2)
so
lim sin (5x) / (2x) , as x --> 0
= lim (5/2)sin (5x)/( (2x)(5/2)
= (5/2) lim sin (5x) / (5x) , as x--->0
= (5/2)(1) = 5/2
try some of the others now after checking on your typing using brackets.
To find the limits algebraically, we will evaluate the given expressions as x (or h or t) approaches the indicated values. Let's go through each limit one by one:
1) lim x->0 sin5x / 2x:
To solve this limit, we can use the property that lim x->0 sin(x) / x = 1.
Using this property, we can rewrite the given expression as 5 * lim x->0 sin5x / (5x).
Now, when x approaches 0, sin5x / (5x) approaches sin(5 * 0) / (5 * 0) = sin(0) / (5 * 0).
Since sin(0) = 0 and any expression divided by 0 is undefined, this limit does not exist.
2) lim x->0 (1 - cosx) / x:
To solve this limit, we will use a trigonometric limit.
Using the limit lim x->0 (1 - cos(x)) / x = 0, we can rewrite the given expression as (1 - cos(x)) / x = (1/x) * (1 - cos(x)).
Now, as x approaches 0, (1/x) approaches infinity (∞), and (1 - cos(x)) approaches 1 - cos(0) = 1 - 1 = 0.
So the limit becomes (∞) * 0, which is an indeterminate form. To evaluate further, we need to use L'Hôpital's rule or other techniques.
3) lim x->7 |x - 7| / (x - 7):
To solve this limit, we will consider the left-hand and right-hand limits separately since the function |x - 7| is not differentiable at x = 7.
- For x < 7, we have |x - 7| = -(x - 7) = 7 - x.
So, when x approaches 7 from the left side, the expression becomes (7 - x) / (x - 7) = -1.
- For x > 7, we have |x - 7| = x - 7.
So, when x approaches 7 from the right side, the expression becomes (x - 7) / (x - 7) = 1.
Since the left-hand and right-hand limits do not match, the limit does not exist.
4) lim x->7 ((x + 2) - 3) / (x - 7):
Simplifying the expression, we get (x - 1) / (x - 7).
As x approaches 7, the expression becomes (7 - 7) / (7 - 7) = 0 / 0, which is an indeterminate form.
To solve this indeterminate form, we can factor out the common factor (x - 7) in both the numerator and denominator: (x - 1) / (x - 7) = 1.
Therefore, the limit is equal to 1.
5) lim h->0 (2 + h)^3 - 8 / h:
Expand (2 + h)^3 using the binomial theorem, we get: 8 + 12h + 6h^2 + h^3 - 8 / h.
Simplifying gives (12h + 6h^2 + h^3) / h = 12 + 6h + h^2.
As h approaches 0, the expression becomes 12 + 0 + 0 = 12.
So, the limit is equal to 12.
6) lim t->0 1/t - 1/t^2 + t:
Simplify the expression by finding a common denominator:
= (t^2 - 1 + t^3)/t^2
As t approaches 0, the expression becomes (0 - 1 + 0)/0^2 = (-1)/0^2 = -1/0.
Dividing any number by zero is undefined, so this limit does not exist.
In summary:
1) The limit does not exist.
2) More work is needed to evaluate the limit.
3) The limit does not exist.
4) The limit is equal to 1.
5) The limit is equal to 12.
6) The limit does not exist.