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Question 5.1
Question 5.1
Solutions
TZ
leumasic
Official
3 months ago
i)
lim
x
→
1
x
2
−
1
x
+
1
=
lim
x
→
1
(
x
−
1
)
(
x
+
1
)
x
+
1
=
lim
x
→
1
(
x
−
1
)
=
0.
\begin{align*} \lim_{x \to 1} \frac{x^2 - 1}{x + 1} &= \lim_{x \to 1} \frac{(x - 1)(x + 1)}{x + 1} \\ &= \lim_{x \to 1} (x - 1) \\ &= 0. \end{align*}
x
→
1
lim
x
+
1
x
2
−
1
=
x
→
1
lim
x
+
1
(
x
−
1
)
(
x
+
1
)
=
x
→
1
lim
(
x
−
1
)
=
0.
ii)
lim
x
→
2
x
3
−
8
x
−
2
=
lim
x
→
2
(
x
−
2
)
(
x
2
+
2
x
+
4
)
x
−
2
=
lim
x
→
2
(
x
2
+
2
x
+
4
)
=
12.
\begin{align*} \lim_{x \to 2} \frac{x^3 - 8}{x - 2} &= \lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{x - 2} \\ &= \lim_{x \to 2} (x^2 + 2x + 4) \\ &= 12. \end{align*}
x
→
2
lim
x
−
2
x
3
−
8
=
x
→
2
lim
x
−
2
(
x
−
2
)
(
x
2
+
2
x
+
4
)
=
x
→
2
lim
(
x
2
+
2
x
+
4
)
=
12.
iii)
lim
x
→
3
x
3
−
8
x
−
2
=
lim
x
→
3
(
x
2
+
2
x
+
4
)
=
19.
\begin{align*} \lim_{x \to 3} \frac{x^3 - 8}{x - 2} &= \lim_{x \to 3} (x^2 + 2x + 4) \\ &= 19. \end{align*}
x
→
3
lim
x
−
2
x
3
−
8
=
x
→
3
lim
(
x
2
+
2
x
+
4
)
=
19.
iv)
lim
x
→
y
x
n
−
y
n
x
−
y
=
lim
x
→
y
(
x
−
y
)
(
x
n
−
1
+
x
n
−
2
y
+
⋯
+
x
y
n
−
2
+
y
n
−
1
)
x
−
y
=
lim
x
→
y
(
x
n
−
1
+
x
n
−
2
y
+
⋯
+
x
y
n
−
2
+
y
n
−
1
)
=
n
y
n
−
1
.
\begin{align*} \lim_{x \to y} \frac{x^n - y^n}{x - y} &= \lim_{x \to y} \frac{(x - y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1})}{x - y} \\ &= \lim_{x \to y} (x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}) \\ &= ny^{n-1}. \end{align*}
x
→
y
lim
x
−
y
x
n
−
y
n
=
x
→
y
lim
x
−
y
(
x
−
y
)
(
x
n
−
1
+
x
n
−
2
y
+
⋯
+
x
y
n
−
2
+
y
n
−
1
)
=
x
→
y
lim
(
x
n
−
1
+
x
n
−
2
y
+
⋯
+
x
y
n
−
2
+
y
n
−
1
)
=
n
y
n
−
1
.
v)
lim
y
→
x
x
n
−
y
n
x
−
y
=
n
x
n
−
1
.
\lim_{y \to x} \frac{x^n - y^n}{x - y} = nx^{n-1}.
y
→
x
lim
x
−
y
x
n
−
y
n
=
n
x
n
−
1
.
vi)
lim
h
→
0
a
+
h
−
a
h
=
lim
h
→
0
a
+
h
−
a
h
(
a
+
h
+
a
)
=
lim
h
→
0
h
h
(
a
+
h
+
a
)
=
lim
h
→
0
1
a
+
h
+
a
=
1
2
a
.
\begin{align*} \lim_{h \to 0} \frac{\sqrt{a+h} - \sqrt{a}}{h} &= \lim_{h \to 0} \frac{a+h-a}{h(\sqrt{a+h}+\sqrt{a})} \\ &= \lim_{h \to 0} \frac{h}{h(\sqrt{a+h}+\sqrt{a})} \\ &= \lim_{h \to 0} \frac{1}{\sqrt{a+h}+\sqrt{a}} \\ &= \frac{1}{2\sqrt{a}}. \end{align*}
h
→
0
lim
h
a
+
h
−
a
=
h
→
0
lim
h
(
a
+
h
+
a
)
a
+
h
−
a
=
h
→
0
lim
h
(
a
+
h
+
a
)
h
=
h
→
0
lim
a
+
h
+
a
1
=
2
a
1
.
0
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