1.) Find the equation of the line that is tangent to the graph of y-y^3 at x=1.
2.) lim x->0 ((sin x*cos 2x)/3x)
3.) Show, using the squeeze theorem, that the limit of Xe^(sin 1/x) as x->0 is 0.
Typo: 1.) Find the equation of the line that is tangent to the graph of y=x^3 at x=1.
1.
y = x^3 , at x = 1, y=1, so we have the point(1,1)
dy/dx = 3x^2 , which is 3, when x = 1
y-1 = 3(x-1)
y-1 = 3x-3
3x - y = 2 or y = 3x - 2
2. Lim sinxcos 2x/(3x)
= lim sinx/x * cos 2x/3, as x ---> 0
= (1)*(1/3) = 1/3
Since |(sin 1/x)| <= 1
|e^(sin 1/x)| <= e
so x e^sin(1/x) <= x * e -> 0 as x->0
1.) To find the equation of the line that is tangent to the graph of y-y^3 at x=1, we can start by finding the derivative of the given function. Let's denote y-y^3 as f(x).
Differentiating f(x) with respect to x, we get:
f'(x) = 1 - 3y^2 * dy/dx
To find the slope of the tangent line at x=1, we need to evaluate f'(x) at x=1. However, we need to find the value of y first.
Since the tangent line is touching the graph of y-y^3 at x=1, we can plug in x=1 into the original equation:
y - y^3 = 1
This gives us a cubic equation, which can be solved to find the values of y. Once we find the values of y, we can substitute them back into f'(x) to find the slope of the tangent line.
After finding the slope, we can use the point-slope form of a line (y - y₁ = m(x - x₁)) to get the equation of the tangent line, where (x₁, y₁) is the point of tangency (in this case, x=1).
2.) To find the limit lim x→0 ((sin x*cos 2x)/3x), we can simplify the expression first.
Let's rewrite the expression as:
lim x→0 (sin x)*(cos 2x)/(3x)
Now, we can use algebraic manipulations and known limits to simplify the expression further.
As x approaches 0, sin x becomes 0 and cos 2x becomes 1. Therefore, the numerator approaches 0.
In the denominator, 3x approaches 0 as x approaches 0.
So, we have:
lim x→0 ((sin x)*(cos 2x)/(3x)) = 0/0
This is an indeterminate form. To evaluate the limit, we can apply L'Hospital's Rule. Taking the derivative of the numerator and denominator, we get:
lim x→0 (cos x)*(-sin 2x + 2 cos 2x) / 3
Now, substituting x=0 into this expression:
= (cos 0)*(-sin 0 + 2 cos 0) / 3
= (1)*(0 + 2*1) / 3
= 2/3
Therefore, the limit of ((sin x*cos 2x)/3x) as x approaches 0 is 2/3.
3.) To show, using the squeeze theorem, that the limit of Xe^(sin 1/x) as x approaches 0 is 0, we need to find two functions whose limits as x approaches 0 are both 0 and sandwich the given function.
First, we can notice that -1 ≤ sin(1/x) ≤ 1 for all x ≠ 0. Hence, multiplying by any positive number (in this case, x), we have -x ≤ x * sin(1/x) ≤ x.
Next, we know that the limit as x approaches 0 of -x is 0 and the limit of x is 0. So, we can use these two functions as the "sandwich" functions.
Therefore, -x ≤ x * sin(1/x) ≤ x
Taking the limit as x approaches 0 for all sides of the inequality, we have:
lim x→0 (-x) ≤ lim x→0 (x * sin(1/x)) ≤ lim x→0 (x)
The left and right limits are both 0 since it is given. Therefore, by the squeeze theorem, the limit of x * sin(1/x) as x approaches 0 is also 0.