- #1
gede
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I have a limit problem. This is the problem:
[tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}[/tex]
The solution is
[tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}[/tex]
[tex]= \lim_{x - 2 \to 0} \frac{\tan [(2 - \sqrt{2x}) × \frac{2 + \sqrt{2x}}{2 + \sqrt{2x}} ]}{x(x - 2)}[/tex]
[tex]= \lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)} [/tex]
Let ##p = x - 2##
and ##2 + \sqrt{2x} = 2 + \sqrt{2⋅2} = 2 + \sqrt{4} = 2 + 2 = 4##
Then
[tex]\lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)} [/tex]
[tex]= \lim_{p \to 0} \frac{\tan [-\frac{1}{2}p]}{2p}[/tex]
[tex]= -\frac{1}{4}[/tex]
Is this correct?
[NOTE: moved to this forum by mentor hence no homework template]
[tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}[/tex]
The solution is
[tex]\lim_{x \to 2} \frac{\tan (2 - \sqrt{2x})}{x^2 - 2x}[/tex]
[tex]= \lim_{x - 2 \to 0} \frac{\tan [(2 - \sqrt{2x}) × \frac{2 + \sqrt{2x}}{2 + \sqrt{2x}} ]}{x(x - 2)}[/tex]
[tex]= \lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)} [/tex]
Let ##p = x - 2##
and ##2 + \sqrt{2x} = 2 + \sqrt{2⋅2} = 2 + \sqrt{4} = 2 + 2 = 4##
Then
[tex]\lim_{x - 2 \to 0} \frac{\tan [ \frac{-2(x - 2)}{2 + \sqrt{2x}} ] }{x(x - 2)} [/tex]
[tex]= \lim_{p \to 0} \frac{\tan [-\frac{1}{2}p]}{2p}[/tex]
[tex]= -\frac{1}{4}[/tex]
Is this correct?
[NOTE: moved to this forum by mentor hence no homework template]
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