Test Your Knowledge: Limits


This page will serve as a practice test with common limit problems and fully worked solutions. At the end, I will review the techniques these problems used, and give a general overview of how to tackle limit problems.


This page will serve as a practice test with common limit problems and fully worked solutions. At the end, I will review the techniques these problems used, and give a general overview of how to tackle limit problems.

1. Solve the limit: lim_{x \to 2} \frac{x^2+x-6}{x-2}

The very first thing we do with limit problems involving rational functions is try to substitute the point in. Since rational functions are continuous in their domain, this could work. Unfortunately for this problem, f(2) is undefined, so we need a different strategy.

Our current problem is that x-2 results in a division by zero when we sub in x =2. Since this is the case, it makes sense for us to try to see if we can remove the x-2, using equivalent functions and cancellation.

To do this, we start with factoring the top function: x^2+x-6 = (x-2)(x+3), which gives us:

lim_{x \to 2} \frac{(x-2)(x+3)}{(x-2)}. From here, we can see that we can cancel (x-2), and remove the undefined portion of the function. Remember that we can do this since it would create an equivalent function for the whole domain.

From here, the problem becomes a simple substitution: lim_{x \to 2} \frac{(x-2)(x+3)}{(x-2)} =  lim_{x \to 2}(x+3) = 5

2. Solve the limit: lim_{x \to 2} \frac{\sqrt{x+2}-3}{x-7}

We can clearly see that x = 7 will create a discontinuity in this function, so let’s see if we can remove the discontinuity. We can’t factor anything here, but we can multiply the top and bottom by the same thing, such that we create an equivalent function without the (x-7).

To do this, we can multiply by \sqrt{x+2}+3. This is known as the conjugate. It will multiply by the top in such a way that the square root will be removed. This will allow us to have more freedom in manipulating the top portion of the rational.

It is important to note that multiplying by this fraction,\frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}, is the same as multiplying by 1, since the top and bottom are the same. This means we are creating an equivelent function, since the top and bottom are modified in a way that is the same as multiplying by 1. After multiplying, we get the following:

lim_{x \to 7} \frac{\sqrt{x+2}-3}{x-7} * \frac{\sqrt{x+2}+3}{\sqrt{x+2}+3}
=lim_{x \to 7} \frac{x+2-3*\sqrt{x+2}+3*\sqrt{x+2}-9}{(x-7)(\sqrt{x+2}+3)}
=lim_{x \to 7} \frac{x+2-9}{(x-7)(\sqrt{x+2}+3)}
=lim_{x \to 7} \frac{(x-7)}{(x-7)(\sqrt{x+2}+3)}

From here, we can see that now we can remove (x-7) which will allow us to use direct substitution to get the answer

=lim_{x \to 7} \frac{(x-7)}{(x-7)(\sqrt{x+2}+3)} = lim_{x \to 7} \frac{1}{\sqrt{x+2}+3} = \frac{1}{6}

3. Show that x^4+x-3 has a root on the interval (1,2)

To show that this function has a root, we can use the intermediate value theorem. The very first thing we need to do with any intermediate value theorem problem is prove that the function we are working with is continous on the interval in question.

In this case, f(x) is a polynomial, so it is continous for any real interval, so we can continue with intermediate value theorem.

Intermediate value theorem tells us that if 0 is between f(1) and f(2), then f(x) must equal 0 at some point in (1,2), meaning it has a root on (1,2).

If we evaluate f(1), we get 1+1-3 = -2. If we evaluate f(2), we get 16+2-3 = 15.

Since 0 is in the interval [-2,15], we can conclude that f(x) has a root in (1,2).

4. Find the limit lim_{x \to \infty}\frac{1}{2*x+3}

When dealing with infinite limits, we want to get an idea of what kind of function we are working with. It becomes very helpful to know about families of functions with these sorts of problems.

In this case, we have a rational function, that has been shifted and scaled. We know that functions in the form \frac{1}{x} will approach 0 as we approach infinity, so we can suspect the same is happening with this function. Look at a graph of this function further confirms this.

We can see that as we approach infinity, the function approaches zero, therefore, lim_{x \to \infty}\frac{1}{2*x+3} = 0

From these problems there are a few key takeaways to consider when solving limit functions.

  1. The very first thing you should try in any limit problem is substituting the value in. If the function is defined at the value, the limit is solved.
  2. If you have a rational function, factor the polynomials to see if the undefined behavior can be removed. Typically, this will be possible to do
  3. If you have a root function somewhere in the function, try multiplying both the function by the conjugate of the root. This will often expose a way to remove the undefined behavior.
  4. With intermediate value theorem questions, it is important to prove that the function is continuous before we apply the theorem. If the function is not continuous on the interval, the theorem doesn’t hold.
  5. With infinite limits, it is often best to try to graph the function, and substitute in large values to see how the function behaves

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