Test Index

NCERT Class XI Mathematics - Limits and Derivatives - Solutions

© examsnet.com
Question : 22 of 72
Marks: +1, -0
lim
x→
π
2
tan2x
x−
π
2
Solution:  
We have,
lim
x→
π
2
tan2x
x−
π
2
 (0/0 form)
Put x =
π
2
 + y; if x →
π
2
 ⇒ y → 0
lim
y→0
tan2(
π
2
+y)
π
2
+y−
π
2
 =
lim
y→0
tan(π+2y)
y
 = (
lim
x→0
tan2y
2y
) × 2 = 2.
© examsnet.com
Go to Question:
Prev Question
Next Question