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2. Examples

1.) Evaluate the following limit:

limx2

x + 2

x2 + 3x + 2

We can do this problem two ways, one using the methods of Chapter 2, and the otherusing LHospitals rule. Well do both ways.

Always we start by plugging in the number:

2 + 2

4 + (6) + 2= 00

So we can try to factor and cancel:

limx2

x + 2

x2 + 3x + 2= lim

x2

x + 2

(x + 2)(x + 1)= lim

x2

1

x + 1=

1

2 + 1= 1

Now we compute the limit again using LHospitals rule.

limx2

x + 2

x2 + 3x + 2

H= lim

x2

(x + 2)

(x2 + 3x + 2)= lim

x2

1

2x + 3=

1

2(2) + 3= 1

2.) Evaluate the following limit:

limx0

x + tanx

sinx

If we plug in 0, we get 00, so we can use LHospitals rule:

limx0

x + tanx

sinxH= lim

x0

(x + tanx)

(sinx)= lim

x0

1 + sec2 x

cosx=

1 + 1

1= 2

3.) Evaluate the following limit:

limx

(lnx)2

x

As x, the top and the bottom , so we have

and we can useLHospitals rule.

limx

(lnx)2

x

H= lim

x

((lnx)2)

(x)= lim

x

2 lnx 1x

1= lim

x

2 ln x

x

Now this limit is simpler than the one we started with, but we still cant evaluate itimmediately because, as x the top and bottom still both go to infinity. We tryusing LHospitals rule again:

limx

2 lnx

x

H= lim

x

(2lnx)

(x)= lim

x

2x

1= lim

x

2

x= 0

So putting it all together,

limx

(lnx)2

x= 0

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5.) Evaluate the following limit:limx0

(cscx cotx)

As x 0, cscx = 1sin x and cotx =cos xsin x

, so we have: . We want torewrite this to get a quotient.

limx0

(cscx cotx) = limx0

1

sin x

cosx

sinx

= lim

x0

1 cosx

sinx

Now notice that as x 0 the top 0 and the bottom 0, so we have: 00 and wecan use LHospitals rule:

limx0

1 cosx

sin x

H= lim

x0

(1 cosx)

(sinx)= lim

x0

sin x

cosx=

0

1= 0

So we can conclude:limx0

(cscx cotx) = 0

Notice that we could have evaluated this limit using the tricks from Chapter 3:

limx0

1 cosx

sinx= lim

x0

1cos xx

sinxx

=0

1= 0

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