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Limits, Continuity, Basic Differentiability, and Computing Derivatives
By: Sameer, Snigdha, Aditya
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Limits Recall that… A limit is when a function gets super close to a number from both sides of x, but the function never reaches that number… It’s predicting a number between two neighboring points. Let’s think about how we can solve these
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Solving Limits - Ex1: Make a table
-0.1 -.01 -.001 0.001 0.01 0.1 Y Plug in on your calculator!
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Solving Limits - Ex1: Make a table (cont.)
-0.1 -.01 -.001 0.001 0.01 0.1 Y 0.998 0.999 1 Therefore, the answer is 1 because as x approaches 0 from both sides, the value of the limit gets closer and closer to 1.
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Solving Limits - Ex2: Graph
As x approaches 0 from both sides, what does f (x) get arbitrarily close to? lim f(x) = 2 x->0
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Solving Limits - Ex3: Direct Substitution
When a function is given, your first tactic to solving a limit should be plugging in the value the limit approaches. Plug in x = 5!
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Solving Limits - Ex4: Factoring/Simplifying
If direct substitution yields an indeterminate answer (0/0 or infinity/infinity), try factoring!
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Solving Limits - Ex4: Factoring/Simplifying
Let’s try factoring.... And now let’s simplify... Now when we use direct substitution, we get 8/2, which is 4
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Solving Limits - Ex5: Rationalizing the Numerator/Denominator
When you can’t use direct substitution or factoring, check for a radical and try rationalizing the numerator or denominator
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Solving Limits - Ex5: Rationalizing the Numerator/Denominator
Multiply either the numerator or denominator by a radical that will get rid of the radical sign and use a different sign for the second term…
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Solving Limits - Ex5: Rationalizing the Numerator/Denominator
Then, simplify:
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Solving Limits - Ex5: Rationalizing the Numerator/Denominator
Finally, use direct substitution:
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Limits - Does Not Exist (DNE)
This limit does not exist because on the limit from the left of x approaches 1 while on the right side of x it approaches 2.
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Limits - Vertical Asymptotes
If f(x) approaches infinity or negative infinity as x approaches c from the right or left, then there is a vertical asymptote at x = c lim f(x) = infinity; therefore VA x->0
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Limits - Vertical Asymptotes (cont.)
A constant divided by a small number yields infinity: This indicates a potential vertical asymptote
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Limits - Horizontal Asymptote
The line y = L is a horizontal tangent if limit as x approaches infinity for f(x) = L
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Limits - Horizontal Asymptotes (cont.)
Use coefficient rule: if highest power in the numerator and denominator match, then divide the coefficients for the limit/horizontal asymptote =3/2 If the power is smaller in numerator or the numerator has a slower function then the limit goes to 0.
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Continuity Consider this hole in the road...is the road continuous?
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Continuity - Definition
Most of you probably agreed that the road is not continuous...you wouldn’t want to drive over that! But...what would make the road continuous? Well, the road would have to: -Fill the hole (discontinuity filled by f(c)) -There would have to be road on either side of the road (limit must exist) -The road on either side and the filled in hole would have to connect (limit must equal f(c)
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Continuity - Definition (cont.)
Similarly in calculus these rules also apply! The limit as x approaches c for f(x) must exist f(c) must exist and have a value The limit as x approaches c for f(x) must equal f(c) Now, justify if each of the points are continuous or not.
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Continuity - Justification
Continuous.The limit as x approaches 0 is 1 and f(0) is also 1 Discontinuous. The limit as x approaches -2 does not exist Discontinuous. The limit as x approaches 3 is 0, but f(3) is -1. Therefore there is a discontinuity at x = 3
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Continuity - Practice Problem
Hmm...let’s use the definition of continuity to solve this problem
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Continuity - Practice Problem (cont.)
First, let’s look at x=1. f(1) = 2 so the limit as x approaches 1 must also equal 2. Use direct substitution!
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Continuity - Practice Problem (cont.)
Plugging in x=1 gives us a + b = 2 Now let’s look at x = 3. f(3) gives -2 so the limit as x approaches 3 must also be -2. Again, use direct substitution… Plugging in x=3 gives us 3a + b = -2
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Continuity - Practice Problem (cont.)
Now, setup a system of equations: a + b = 2 3a + b = -2 Solving for a gives - 2a = 4 a = -2. Plugging back in, we get -2 + b = 2, and b = 4
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Basic Differentiability
A function is differentiable if its derivative exists at every point on its domain. A derivative is essentially the slope of a function and is denoted as f’(x) Differentiability implies Continuity However, continuity does not mean a function is 100% differentiable
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Basic Differentiability - Non-differentiable functions
Corner Cusp
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Basic Differentiability - Non-differentiable functions (cont.)
Vertical Tangent Jump Discontinuity
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Computing a Derivative
Formal Definition of a Derivative This simply means the derivative of cos(x) at x= pi
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Computing a Derivative - Polynomials
Here’s the formula for computing a derivative* *Note: there are special ways to derive trig, logarithmic, e^x, and other special case functions
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Computing a Derivative - Polynomials (cont.)
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Computing a Derivative - Rational Functions
Quotient Rule: When computing functions with quotients, you must use quotient rule.
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Computing a Derivative - Rational Functions (cont.)
Use quotient rule!
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Computing a Derivative - Rational Functions (cont.)
Derivative of f(x)!
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Computing a Derivative - Trigonometric Functions
Memorize these Trig Derivatives!
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Computing a Derivative - Trigonometric Functions (cont.)
Quick introduction to chain rule: a derivative rule in which nested functions must be included in the derivative.
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Computing a Derivative - Trigonometric Functions (cont.)
h(x) = sin(2x)...Find the derivative f(x) = sin(x) g(x) = 2x so: h’(x) = cos(2x)*2 or 2cos(2x)
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Computing a Derivative - Exponential/Logarithmic Functions
You should know these derivatives:
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Computing a Derivative - Exponential/Logarithmic Functions (cont.)
Finally, product rule:`
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Computing a Derivative - Exponential/Logarithmic Functions (cont.)
Let’s take the derivative of this! Using chain rule, we get:
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That’s All Folks… FIN.
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