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AP Calculus AB / BC Big Idea Review - Limits

Introduction

A limit is the value that a function takes as it approaches a certain point. For continuous functions, this is the value of the function itself. However, for discontinuous functions, limits can have interesting implications. In other words, if the function is continuous:

$$ \lim_{x \to a} f(x) = f(a)$$

When a function is not continuous is when interesting things start happening with Limits. In order to understand continuity, we must first understand discontinuity and then derive its opposite property. There are many interesting types of discontinuities that can occur, which we won't go into here, but let's take an example to understand this.

Limits of discontinuous functions

Let's take a piecewise function, f(x), made up of 3 continuous functions such that f(x) is equal to:

1) $$x \text{ if } x < 0$$
2) $$ x^2 \text{ if } 0\leq x \leq 3 $$
3) $$2x + 2 \text{ if } x > 3$$

Find the limit of f(x) as x approaches 3 from the right and from the left. Take a moment and find these two limits. The answer is below.

Answer: 8 from the right, 9 from the left

When x approaches 3 from the right (decreasing-x) direction, the third piecewise function is acting on f(x), meaning f(x) = ( 2 • 3 ) + 2 = 8. When x approaches 3 from the left, (increasing-x) f(x) = x2 = (3)2 = 9.

Based on this information, we can tell that there is a discontinuity at x = 3 because the left hand side limit and the right hand side limit don't equal each other. This isn't always the case with piecewise functions. The above function is continuous at x = 0, for example.

Squeeze theorem

For some functions, it is difficult to evaluate their limits analytically, especially limits at infinity. This is where the Squeeze Theorem sometimes comes in handy. The Squeeze Theorem says that if a function can be bounded by an upper bound function and a lower bound function, and the limits of each of those function is known and converges to the same value, then the limit of the bound function also converges to that value. We can use the Squeeze Theorem to derive the limit

$$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $$

and

$$ \lim_{x \to 0} {\frac{\cos x - 1}{x}} = 0 $$

L’Hospital’s Rule

There's another way to prove the limits above using derivatives to solve limits of indeterminate forms. Indeterminate forms usually occur with limits at 0 or infinity, and occur if a limit evaluates to one of the following:

$$ 0 \cdot \infty $$ $$ -\frac\infty\infty $$ $$ \frac\infty\infty $$ $$ \frac 0 0 $$ $$ 0\cdot-\infty $$ $$ 0^0 $$ $$ 1 ^ \infty $$

You don't have to remember all of them, you'll know it when you see that the limit is unsolvable. At that point it's time to rearrange the limit and use L'Hospital's Rule, which states that for any indeterminate form:

$$ \lim_{x \to a} \frac{f(x)}{g(x)} $$

$$ = \lim_{x\to a} \frac{f'(x)}{g'(x)}$$

So to prove the famous trigonometric limit in the example from the Squeeze Theorem, we take the limit equal to the derivative of the numerator divided by the derivative of the denominator:

$$ \lim_{x \to 0} \frac{\sin x}{x} $$

$$ = \lim_{x \to 0} \frac{\cos x}{1} $$

$$ = \frac{1}{1} = 1 $$

To view more practice questions like the ones discussed in this article, get the new Omninox practice questions eBook for all your devices from Amazon by pressing on the book cover above.

Practice Problems

Example 1

Find $$ \lim_{x\to0}x^x $$

  1. 0
  2. 1
  3. e
  4. None of the Above

Answer: B

Solution: The trickiest part here is rewriting the limit as

$$ \lim_{x\to0}e^{\ln x^x} $$

$$ =\lim_{x\to0}e^{x\cdot\ln x} $$

Now we’re stuck. We don’t know what the exponent equals as it approaches 0, so let’s prove its limit separately to figure it out.

$$ \lim_{x\to0}x\ln x $$

$$ =\lim_{x\to0}\frac{\ln x}{\frac1x} $$

We can rewrite it this way for the same reason that 2*4=2/(1/4), just replace 4 with x, and 2 with ln(x). Now we can apply L’Hospital’s Rule to the above limit:

$$ \lim_{x\to 0}\frac{\ln x}{\frac 1 x}$$

$$ =\lim_{x\to 0}\frac{\frac{x}{dx}\ln x}{\frac{d}{dx}\frac{1}{x}} $$

$$ =\lim_{x\to 0}\frac{\frac 1 x}{-\frac{1}{x^2}} $$

$$ =\lim_{x\to 0} -\frac{x^2}{x} $$

$$ = \lim_{x \to 0}-x $$

Which equals 0 as x approaches 0. We can plug this limit back into our original limit, to get:

$$ \lim_{x\to0}e^{x\cdot\ln x} $$

$$ =e^{\lim_{x\to0}x\cdot \ln x} $$

$$ =e^0=1 $$

Remember that this was just another way to rewrite our original limit. So that is how we know that

$$ \lim_{x\to 0}x^x=1 $$

Example 2

Evaluate the following limit

$$ \lim_{x\rightarrow \infty}4\cdot \sin(\ln(x)) $$

  1. 0
  2. 4
  3. -4
  4. The limit does not exist

Answer: D

Solution: As x approaches infinity in the above function, although sin(x) is a bounded function, ln(x) continues to increase to infinity and the function oscillates between 4 and -4 faster and faster to the point where it actually becomes discontinuous. This is a special type of discontinuity called a compression discontinuity or an oscillation discontinuity.

Example 4

Find the limit

$$ \lim_{x\to-3}\frac{\sin\text{ }(x^2-9)}{x+3} $$

  1. 0
  2. -3
  3. 3
  4. -6

Answer: D

Solution: This limit looks similar to

$$ \lim_{x \to 0}\frac{\sin x}{x}=1 $$

but the numerator and the denominator do not match. How can we fix this? Recall that one of the techniques we can use to rewrite limits that look like this is to multiply by the conjugate of the limit. This time we are multiplying so that the denominator can become a difference of two squares. This way,

$$ \lim_{x\to-3}\frac{\sin\text{ }(x^2-9)}{x+3} $$

becomes

$$ =\lim_{x\to-3}\frac{\sin(x^2-9)}{x+3}\cdot\frac{x-3}{x-3} $$

$$ =\lim_{x\to-3}\frac{(x-3)\cdot\sin(x^2-9)}{x^2-9} $$

$$ =\lim_{x\to -3} (x-3) $$

$$ \cdot \lim_{x\to-3} \frac{\sin(x^2-9)}{x^2-9} $$

We know what the second limit equals from the formula, so we can rewrite the limit as

$$ =\lim_{x\to-3}(x-3) \cdot 1 $$

$$ = -6 $$

Example 5

Solve for

$$ \lim_{x\to\infty}\frac{x+2}{3x+4} $$

  1. 0
  2. 1/3
  3. 1
  4. Limit DNE

Answer: B

Solution: This is a special-case of limits that approach -∞ or ∞. There is a shortcut to evaluating these limits when the numerator is a polynomial and so is the denominator, and the highest degree of the exponent is the same for both the numerator and denominator. In this case the highest degree is 1.

The limit of this special case is the coefficient of the numerator's highest degree monomial divided by the coefficient of the denominator's highest degree monomial.

That's a fancy way of saying: ignore everything except the highest-degree monomial (1x in the numerator, 3x in the denominator). Then divide the coefficients to get 1/3.

Proving this isn't too difficult. Conceptually, it has to do with how "quickly" each term "reaches" infinity. Remember that ∞ plus any constant is still ∞ (which means we can ignore the constants 2 and 4 in the above problem). You can continue to do that and factor out each "x" out of like terms for higher order polynomials.

This also has the implication of another special case limit approaching ∞:

$$ \lim_{x\to\infty} \frac{x ^ n}{a^x} $$

Where n and a are both constants. This limit will always equal 0 and its inverse will always be infinity. This is because any exponentially increasing function, whether it's ex or 2x or 10x, will continue to "reach" infinity at an ever-accelerating rate. You don't have to understand exactly why this is true unless you're studying a more advanced Calculus than AP Calc AB, but you can check that it's true graphically using any graphing software (just be sure to zoom out enough to see the trend of both lines at a large value of x.

To view more practice questions like the ones discussed in this article, get the new Omninox practice questions eBook for all your devices from Amazon by clicking the book cover above.

Conclusion

In addition to the practice questions discussed above, there are a few other techniques and types of questions you might see for solving limits. Almost all of them boil down to finding a way to cancel out one of the terms. Algebraically rearrange the function given until you can cancel something out. That is when you know you are approaching the solution.