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Byju's Answer
Standard XII
Mathematics
Existence of Limit
Evaluate : ...
Question
Evaluate :
lim
x
−
1
x
7
−
2
x
5
+
1
x
3
−
3
x
2
+
2
Open in App
Solution
lim
x
→
1
x
7
−
2
x
5
+
1
x
3
−
3
x
2
+
2
on differentiating both numertor & denomitator w.r.t x we get
⇒
lim
x
→
1
d
d
x
(
x
7
−
2
x
5
+
1
)
d
d
x
(
x
3
−
3
x
2
+
2
)
⇒
lim
x
→
1
7
x
6
−
10
x
4
3
x
2
−
6
x
on putting limit we get
⇒
7
(
1
)
6
−
10
(
1
)
4
3
(
1
)
2
−
6
(
1
)
7
−
1
3
−
6
=
−
3
−
3
=
1
Hence
lim
x
→
1
x
7
−
2
x
5
+
1
x
3
−
3
x
2
+
2
=
1
Hence
lim
x
→
1
x
7
−
2
x
5
1
x
3
−
3
x
2
+
2
=
1
.
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1
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