Calculus 1 with the Math Sorcerer

Use the squeeze theorem to bound x^2 cos(1/x) between -x^2 and x^2 as x approaches zero, proving the limit is zero.

Apply the squeeze theorem to x^4 cos(17/x) as x approaches zero, using -1 ≤ cos(17/x) ≤ 1 to bound -x^4 ≤ x^4 cos(17/x) ≤ x^4, hence the limit is zero.

Learn to prove that a bounded function times a function tending to zero has a zero limit using the squeeze theorem, by bounding g(x) and applying absolute values.

Evaluate the limit as x approaches zero of sin(3x)/(2x) by rewriting it as (3/2)·(sin(3x)/(3x)) and using the limit equals one, yielding 3/2.

This lecture computes the limit of sin(4x)/x as x approaches zero, using sin t / t = 1 and the substitution t = 4x to obtain 4.

Apply the standard limit sin x over x equals one to evaluate limits of sin(3x) and sin(2x) as x approaches zero, yielding a result of three halves.

Compute the limit of tan x over x as x approaches zero using two methods: rewrite with sine and cosine and standard limits, or apply L'Hôpital's rule to get 1.

This lecture demonstrates evaluating a trig limit by rewriting with sin(7x)/(7x) and sin(3x)/(3x), canceling factors, and applying the standard limit to finish.

Explore limits with sine expressions by transforming them, using a difference of squares, and forming a product of limits to evaluate a sine-based limit as x approaches one.

Learn to evaluate a trig function limit as x approaches zero by multiplying by the reciprocal to simplify. Use sine, cosine, and tangent relationships to substitute and obtain zero.

Learn to prove the limit of f(x)=3x+5 as x approaches 2 equals 11 using the delta-epsilon definition, including scratch work and a formal proof.

Prove that the limit of x^2 minus 4 as x approaches 2 is 4 using the delta-epsilon definition. The lecture uses factoring, bounding |x+2| via |x-2|<1, and sets delta = min(1, epsilon/5) to finish.

Define continuity as drawing a function without lifting your pen, requiring three conditions: defined at c, the limit exists, and the limit equals f(c); the third implies the first two.

Explore removable vs non removable discontinuities, and see how cancellation can remove a discontinuity and how limits indicate removability, with holes and vertical asymptotes in rational functions.

Learn how removable and nonremovable discontinuities arise in a rational function by factoring, canceling common factors to reveal holes, and identifying vertical asymptotes in Example 1.

Identify discontinuities by zeros of the denominator, with x = k pi for sin x and x = pi/2 + k pi for cos x, yielding vertical asymptotes that are nonremovable.

Identify all discontinuities of the given trig function and classify them as non-removable vertical asymptotes at x = 1 + 2k (k integer), by solving cos(pi x/2) = 0.

Explore making a piecewise function continuous by matching left and right limits at negative one and at three, solving for A and B to connect the segments.

Determine c values that make the piecewise function continuous on the real line by equating left and right limits, solving the equation c^2 + c - 1 = 0.

Explains the intermediate value theorem, showing that a continuous function on a closed interval takes every value between f(a) and f(b) and finds a c in (a,b) with f(c)=n.

Explore infinite limits, where f(x) grows without bound as x approaches c, with left and right behaviors and vertical asymptotes.

Examine the limit as x approaches three from the right of one over x minus three, showing the expression diverges to infinity.

Explore right-hand limits of a rational function as x approaches two from the right; observe how a positive numerator over a shrinking positive denominator drives the quotient to infinity.

Analyze infinite limit of x over cos x as x approaches pi/2 from the right. Cos x approaches zero from the negative side, so the ratio diverges to negative infinity.

Explore vertical asymptotes through limits, identifying x=c where f(x) approaches infinity or negative infinity from left, right, or both, and distinguish removable from non removable discontinuities.

Show that the limit as x approaches -1 of sin(x+1)/(x+1) equals 1, yielding a removable discontinuity. There is no vertical asymptote at x = -1.

Identify where tan(3x) is undefined by setting cos(3x)=0; solve for x: x = pi/6 + k pi/3, where k is any integer, giving the vertical asymptotes.