Evaluate limx→2x10−1024x5−32
64
We have, limx→2x10−1024x5−32
At x=2, the expression x10−1024x5−32 is the form 00.It is an indeterminant form. Already, we have solved these types of problems using standard limits.
Now, we will solve it with a ver effective method called L'hospital rule.
here, if we have indeterminate form of 00 . so, we need to do is differentiate the numerator and differentaiate the denominator and take the limit.
= limx→2ddx(x10−1024)ddx(x5−32)
= limx→210x9−05x4−0 {ddxxn=nxn−1}
= limx→22x5=2×25=64