document.write( "Question 1204215: If limt(e^(ln(ax)) × tan(2a/x))= 8 ' find value of a ? \n" ); document.write( "

Algebra.Com's Answer #840326 by ikleyn(52786) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "        In the post by @MathLover1, this woman made very rude mistakes, which show that
\n" ); document.write( "        she not only does not know Calculus, but does not know the logarithmic function, as well.\r
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\n" ); document.write( "\n" ); document.write( "        Therefore, I will explain the solution from the very beginning.\r
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document.write( "If the problem asks about the limit  at x---> -oo,  then it is clear that the coefficient \"a\"\r\n" );
document.write( "in this consideration must be negative - otherwise, logarithm  ln(ax)  is NOT DEFINED.\r\n" );
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document.write( "With negative \"a\", ln(ax) is defined at negative x, and  we can write   = ax.\r\n" );
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document.write( "Then  \r\n" );
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document.write( "     =  =   = 8,\r\n" );
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document.write( "which implies  = 4,  and since \"a\" is negative,  a = -2.\r\n" );
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document.write( "So, a = -2 is the only solution to this problem at x ---> -oo, which is exactly opposite to the conclusion by @MathLover1.\r\n" );
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document.write( "If the question is about finding \"a\" from this equation at x ---> oo,  then  the answer is  a = 2.\r\n" );
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document.write( "Thus, the ANSWER is twofold: if x ---> -oo,  then a = -2.\r\n" );
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document.write( "                             if x --->  oo,   then a = 2.\r\n" );
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\n" ); document.write( "\n" ); document.write( "How this woman can call herself \" MathLover1 \", making such errors, is a mystery to me.\r
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