Functions Question 215
Question: $ \underset{x\to 4}{\mathop{\lim }},
[ \frac{{x^{3/2}}-8}{x-4} ]= $ [DCE 1999]
Options:
A) 3/2
B) 3
C) 2/3
D) 1/3
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=\underset{x\to 4}{\mathop{\lim }},[ \frac{{x^{3/2}}-8}{x-4} ] $ $ =\underset{x\to 4}{\mathop{\lim }}[ \frac{{{({x^{1/2}})}^{3}}-{{(2)}^{3}}}{(\sqrt{x}-2)(\sqrt{x}+2)} ] $
Þ $ y=\underset{x\to 4}{\mathop{\lim }},\frac{({x^{1/2}}-2)(x+4+2\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)} $
Þ $ y=\underset{x\to 4}{\mathop{\lim }},\frac{(x+4+2\sqrt{x})}{(\sqrt{x}+2)} $ $ =\frac{4+4+2\sqrt{4}}{\sqrt{4}+2} $ $ =\frac{12}{4}=3 $ . Trick : Applying L-Hospital?s rule, we get $ \underset{x\to 4}{\mathop{\lim }},\frac{\frac{3}{2}{x^{1/2}}}{1} $ $ =\frac{3}{2}{{(4)}^{1/2}} $ = 3.
BETA
