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Question 5.2
Question 5.2
Solutions
TZ
leumasic
Official
3 months ago
i)
lim
x
→
1
1
−
x
1
−
x
=
lim
x
→
1
1
−
x
(
1
−
x
)
(
1
+
x
)
=
lim
x
→
1
1
1
+
x
=
1
2
.
\begin{align*} \lim_{x \to 1} \frac{1-\sqrt{x}}{1-x} &= \lim_{x \to 1} \frac{1-x}{(1-x)(1+\sqrt{x})} \\ &= \lim_{x \to 1} \frac{1}{1+\sqrt{x}} \\ &= \frac{1}{2}. \end{align*}
x
→
1
lim
1
−
x
1
−
x
=
x
→
1
lim
(
1
−
x
)
(
1
+
x
)
1
−
x
=
x
→
1
lim
1
+
x
1
=
2
1
.
ii)
lim
x
→
0
1
−
1
−
x
2
x
=
lim
x
→
0
1
−
(
1
−
x
2
)
x
(
1
+
1
−
x
2
)
=
lim
x
→
0
x
2
x
(
1
+
1
−
x
2
)
=
lim
x
→
0
x
1
+
1
−
x
2
=
0.
\begin{align*} \lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{x} &= \lim_{x \to 0} \frac{1-(1-x^2)}{x(1+\sqrt{1-x^2})} \\ &= \lim_{x \to 0} \frac{x^2}{x(1+\sqrt{1-x^2})} \\ &= \lim_{x \to 0} \frac{x}{1+\sqrt{1-x^2}} \\ &= 0. \end{align*}
x
→
0
lim
x
1
−
1
−
x
2
=
x
→
0
lim
x
(
1
+
1
−
x
2
)
1
−
(
1
−
x
2
)
=
x
→
0
lim
x
(
1
+
1
−
x
2
)
x
2
=
x
→
0
lim
1
+
1
−
x
2
x
=
0.
iii)
lim
x
→
0
1
−
1
−
x
2
x
2
=
lim
x
→
0
1
1
+
1
−
x
2
=
1
2
.
\begin{align*} \lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{x^2} &= \lim_{x \to 0} \frac{1}{1+\sqrt{1-x^2}} \\ &= \frac{1}{2}. \end{align*}
x
→
0
lim
x
2
1
−
1
−
x
2
=
x
→
0
lim
1
+
1
−
x
2
1
=
2
1
.
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