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Indeterminate forms of Limits
Examples with detailed solutions and exercises that solves limits questions related to indeterminate forms such as :
Theorem
A second version of L'Hopital's rule allows us to replace the limit problem
with another simpler problem to solve.
Examples with Solutions
Example 1
Find the limit
Solution to Example 1:
Since
and
we have the indeterminate form
The above L'Hopital's rule can be used to evaluate the given limit question as follows
Evaluate the derivatives in the numerator and the denominator
Evaluate the limits in the numerator and denominator
We now evaluate the given limit
Example 2
Find limx→∞xe−x
Solution to Example 2:
Note that
limx→∞x=∞
and
limx→∞e−x=0
This is the indeterminate form ∞⋅0. The idea is to convert it into to the indeterminate form ∞∞ and use L'Hopital's theorem. Note that
limx→∞xe−x=limx→∞xex=∞∞
We apply the above L'Hopital's theorem
=limx→∞(x)′(ex)′=limx→∞1ex=0
Example 3
Find limx→∞(1+1/x)x
Solution to Example 3:
Note that limx→∞(1+1/x)=1 and the above limit is given by
limx→∞(1+1/x)x=1∞
which is of the Indeterminate form 1∞. If we let t=1/x the above limit may written as
limx→∞(1+1/x)x=limt→0(1+t)1/t , note that as x→∞ , t→0
Let y=(1+t)1/t and find the limit of lny as t approaches 0
lny=ln(1+t)1/t=(1/t)ln(1+t)
The advantage of using lny is that
limt→0lny=limt→0ln(1+t)t=00
and the limit has the indeterminate form 0 / 0 and the first L'Hopital's rule can be applied
limt→0lny=limt→0ln(1+t)t
=limt→0(ln(1+t))′(t)′=limt→01/(1+t)1=1
Since the limit of ln y = 1 the limit of y is e 1 = e, hence
limx→∞(1+1/x)x=e
Example 4
Find the limit limx→0+(1/x−1/sinx)
Solution to Example 4:
Note that
limx→0+(1/x)=+∞
and
limx→0+(1/sinx)=+∞
This limit has the indeterminate form ∞−∞ and has to be converted to another form by combining 1/x−1/sinx
limx→0+(1/x−1/sinx)=limx→0+sinx−xxsinx=00
We now have the indeterminate form 0 / 0 and we can use the L'Hopital's theorem.
limx→0+sinx−xxsinx=limx→0+(sinx−x)′(xsinx)′=limx→0+cosx−1sinx+xcosx=00
We have again the indeterminate form 0 / 0 and use the L'Hopital's theorem one more time.
=limx→0+(cosx−1)′(sinx+xcosx)′
=limx→0+−sinxcosx+cosx−xsinx=02=0
Example 5
Find the limit limx→0+xx
Solution to Example 5:
We have the indeterminate form 00. Let y = x x and ln y = ln (x x) = x ln x. Let us now find the limit of ln y
limx→0+lny=limx→0+xlnx=0⋅∞
The above limit has the indeterminate form 0⋅∞. We have convert it as follows
limx→0+xlnx=limx→0+lnx1/x=∞∞
It now has the indeterminate form ∞∞ and we can use the L'Hopital's theorem
limx→0+lnx1/x=limx→0+(lnx)′(1/x)′
=limx→0+1/x−1/x2
=limx→0+−x=0
The limit of ln y = 0 and the limit of y = xx is equal to
e0=1
Exercises
Find the limits
1. limx→∞(lnx)1/x
2. limx→∞(lnx−ln(1+x))
3. limx→∞xex
4. limx→0+xsinx
Solutions to Above Exercises
1. 1
2. 0
3. 0
4. 1
More Links on Limits
Calculus Tutorials and Problems
Limits of Absolute Value Functions Questions