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Question 15 - Week 8 Practice | Prasnya
Q15
00:00
4 Aug 2024
Q15.
Calculate the limit of the following function at
x
=
1
x=1
x
=
1
:
f
(
x
)
=
x
4
−
3
x
3
+
2
x
4
−
5
x
3
+
3
x
2
+
1
f(x)=\frac{x^4-3x^3+2}{x^4-5x^3+3x^2+1}
f
(
x
)
=
x
4
−
5
x
3
+
3
x
2
+
1
x
4
−
3
x
3
+
2
, if
x
≠
1
x\ne1
x
=
1
.
@passage
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Q15
00:00
4 Aug 2024
Q15.
Calculate the limit of the following function at
x
=
1
x=1
x
=
1
:
f
(
x
)
=
x
4
−
3
x
3
+
2
x
4
−
5
x
3
+
3
x
2
+
1
f(x)=\frac{x^4-3x^3+2}{x^4-5x^3+3x^2+1}
f
(
x
)
=
x
4
−
5
x
3
+
3
x
2
+
1
x
4
−
3
x
3
+
2
, if
x
≠
1
x\ne1
x
=
1
.
@passage
Your Answer
Save
Read
Check
Details
