Evaluating limits

Some formulas are difficult to evaluate their limits, however, using L’Hospital’s Rule which says that with functions in the form of. 0/0 , (infinity)/(infinity) 0^0, 1^infinity.
We can simply simplify the quotient and take the derivative until our value will work in the remaining function.

Lets take a look at some examples:

Example 1: Suppose we want to take the limit of (x^2-16)/(x-4) as x approaches 4.
If we simply plug in 4 into the function we get (4^2-16)/(4-4) = 0/0
So L’Hospital’s Rule says that we can take the derivative d/dx of the numerator and the derivative of the denominator so lets try that:
Taking d/dx(x^2-16)/d/dx(x-4) we get (2x)/1
Now we may plug in 4 into our new function to get that the limit as x approaches 4 or (x^2-16)/(x-4) = (2)*(4)/1 = 8

Lets take a look at another example:

Example 2: Suppose we want to take the limit of (4x^2-5x)/(1-3x^2) as x approaches infinity
The problem here is that infinity is not a number therefore we can not plug it into the original formula. So lets try take the derivative of the numerator and the derivative of the denominator to see what we yield.
d/dx(4x^2-5x)/d/dx(1-3x^2) = (8x-5)/(6x) we still cannot plug anything into this equations so we will again take the derivate of the numerator and the denominator of
(8x-5)/(6x), lets see d/dx(8x)/d/dx(6x) = 8/6 = 4/3 giving us that the limit as x approaches infinity of (4x^2-5x)/(1-3x^2) is 4/3

Lets do more

Example 3: Suppose we are ask to find the limit of sin(x)/x as x approaches 0 if we plugged 0 into sin(x)/x we get 0/0. so lets take the derivative of the numerator and the denominator of sin(x)/x Doing so we get d/dx(sin(x))/d/dx(x) = cos(x)/1
Now we may plug in 0 into cos(x)/1 to cos(0)/1 = 1/1
So the limits as x approaches 0 of sin(x)/x = 1

Lets try a trickier one:

Example 4: Suppose that we want to find the limit of e^x/x^2 as x approaches affinity. If x is approaching positive infinity then we have that the numerator and the denominator are both approaching affinity. Lets try to apply L’Hospital’s Rule and take the derivative of the numerator and of the denominator and see what happens.
Taking d/dx(e^x)/d/dx(x^2) we get (e^x)/(2x) so this tell us that e^x is increasing faster than 2x so this limit should be infinity, lets check by applying L’Hospital’s Rule again to (e^x)/(2x)
We get that d/dx(e^x)/d/dx(2x) = e^x/2 Now we can see that as e^x increases then 2 will go into it more times to give us the limit as x approaches infinity of e^x/x^2 = infinity

Example 5: Suppose that we want to find the limit of (5t^4-4t^2-1)/(10-t-9t^3) as x approaches 1
More numbers but don’t worry with what we know those will all go away.
Lets apply L’Hospital’s Rule and take the derivative of the numerator and the dominator multiple times to get that d/dx(5t^4-4t^2-1)/d/dx(10-t-9t^3) = (20t^3-8t)/(-1-27t^2)
we could plug in 1 into (20t^3-8t)/(-1-27t^2)
To get 20-8/(-1-27) = 12/(-28) = -6/14 = -3/7 so the limit of (5t^4-4t^2-1)/(10-t-9t^3) as x approaches 1 = -3/7

Lets try applying L’Hospital’s Rule other types of limits:

Example 6: Suppose that we want to find the limit of xlnx as x approaches 0 This function is not in quotient form but we can put it in quotient form xlnx = (lnx)/(1/x)
Now lets take the derivative the numerator and the denominator to get
d/dx(lnx)/d/dx(1/x) = (1/x)/((x*0)-1(1)/(x^2)) simplifying we get (1/x)/(-1/x^2) = x^2/-x = (-x) now we can plug in 0 into (-x) to get (-0) = 0
Therefore the limit as x approaches 0 of xlnx = 0

Lets try this one:

Example 7: Find the limit of xe^x as x approaches negative infinity
Since this is not in quotient form we can apply L’Hospital’s Rule until it is, so lets put it in quotient form. xe^x = (e^x)/(1/x) or xe^x = (x)/(1/e^x)
Now lets apply L’Hospital’s Rule on (e^x)/(1/x) d/dx(e^x)/d/dx(1/x) = (e^x)/(-1/x^2) so that is not going to help us. What if we apply L’Hospital’s Rule to (x)/(1/e^x) then we get d/dx(x)/d/dx(1/e^x) = 1/(e^x*0)-(e^x*1)/(e^x)^2 = (1)/(-e^x)/(e^x)^2 simplifying we get 1/(-e^x) therefore as x approaches –infinity the denominator is getting bigger and which means that 1/(-e^x) will go to 0, so the limit as x approaches negative infinity
of xe^x = 0

Ok one more

Example 8: This one is tricky! Find the limit of x^(1/x) as x approaches infinity it looks like it will be zero but don’t let the eyes deceive you because as x if x = infinity then (1/x) will go to zero witch means that we will have (infinity)^0 which any number x^0 = 1 so the limit of x^(1/x) should be 1
Lets check:
Let y = x(1/x)
Using our friend (ln) we can attach natural log (ln) to both side of the equation to give us
lny = lnx^(1/x) so lny = 1/x(lnx)
now we can see that as x get larger then 1/x will go to 0 which would make 1/x(lnx) = 0
however because we attached natural log to y we must get rid of the natural log by raising the equation lny = 1/x(lnx) by (e) so lets do that. Since we found that as x goes to infinity in 1/x(lnx) then 1/x(lnx) = 0 we can say e^(lny) = e^0 giving us y = 1
Therefore the limit as x approaches infinity of x^(1/x) = 1