# Calculus Test 1

(Saudi)

Here's a calculus midterm sent by Fahad. If you can solve it would be great if you posted your solution. If you want to do so, you can create your own page below.

Calculus Midterm

Below is my solution.

Ok, the fi rst problem is to solve some limits. The first one is the quotient between two polynomials: As we saw at the page about limits at infi nity, in this case we must divide both the numerator and denominator by x to the greatest power that appears in the expression: The second one is a lateral limit: We note that x is always smaller than 4, so all three factors there are always negative. Then we can separate the limit: The third limit is also a lateral limit (there's a typo in the original pdf, the value to which x approaches must be 2 instead of x): We note again that both the numerator and denominator are always negative, so: Then we have a trigonometric limit: There's a very similar limit at the page about the squeeze theorem. Here we multiply and divide by 3: The second problem is about continuity. We're asked to find the constant k that will make the following function coninuous: Here, the only "problem" point is x=4. In all other points the function is necessarily continuous, because it is a polynomial function, which is continuous.

To make it continuous at point x=4, from the de nition of continuity the following condition must be met: That is: That limit is easy to calculate: And from that equation we get: The third question is about asymptotes. I haven't written about asymptotes in my site yet, but this one is easy to understand. Asymptotes can be vertical and horizontal. They're basically lines in the plane to which the function approaches as they tend to infi nity.

We can either have the function tending to infi nity or the independent variable x approaching infi nity.

In the fi rst case we'll have a vertical asymptote and in the second a horizontal one.

So, let's find fi rst the vertical asymptotes. We must find the values at which thefunction becomes "infinity". Our function is: We have x-1 in the denominator. So, our function will approach in nity as
x approaches 1. That means that we have a vertical asymptote at x=1.

To fi nd if we have horizontal asymptotes, let's calculate the limits: We can see that: The same is true for the second limit. So, we have a horizontal asymptote at y=2.

The last problem asks us to find, using the defi nition, the derivative of the function: Applying the definition: At x=2, we have f'(2) = 12.

### Comments for Calculus Test 1

Average Rating     Mar 11, 2012 Rating     thank you so much by: Anonymous Thank you you are same my big brother i love you all amrican

 Mar 10, 2012 Rating     There is an Error by: Pablo There is an error there in my solution. Can you spot it?