Finding the limit

Limit as x approaches -1 of (x^2-1)/(x+1)
Solution: if x = -1 then the denominator would equal 0 which is undefined.
Therefore we are interested in finding out what the follow of (x^2-1)/(x+1) is as x gets close to -1
We can do this by taking the derivative of both the top and bottom, then plug in -1 and evaluate.

d/dx (x^2-1)/d/dx (x+1) = 2x/1; therefore the limit as x approaches -1 of (x^2-1)/(x+1) is 2(-1)/1 = -2

As illustrated in the graph below when x gets closer to -1 then the value of (x^2-1)/(x+1) gets closer to -2 therefore we know that (x^2-1)/(x+1) is always increasing.



Limit as x approaches 1 of (x^9-1)/(x^5-1)
Solution:
Step one: find the d/dx(x^9-1)/d/dx(x^5-1) = 9x^8/5x^4
Step two: evaluate 9x^8/5x^4 when x =1 =9/5
Therefore the limit as x approaches 1 of (x^9-1)/(x^5-1) = 9/5 = 1.80

As illustrated in the graph below when x gets closer to -1 then the value
Of (x^9-1)/(x^5-1) gets closer to 1 the value of (x^9-1)/(x^5-1) gets closer to 1.80




Limit as t approaches 0 of (e^t-1)/(t^3)
Solution:
Step one:
This one is a little more trickier and requires more than the previous ones because if we find d/dx(e^t-1)/d/dx(t^3) we get e^t/3t^2 and we still can’t evaluate x = 0 because the donator will still = 0 and that is undefined so we proceed with the following:
However we can see that if as t gets closer to 0 e^t gets closer to 1 and as t gets closer to 0 then 3t^2 gets closer to 0 therefore e^t/3t^2 goes to infinity and so the Limit as t approaches 0 of (e^t-1)/(t^3) is infinity as illustrated in the graph below:

Graph of t approaches 0 from the positive side



Graph of t approaches 0 from the from the negitive side