document.write( "Question 1158629: Suppose f(x) is defined as shown below
\n" ); document.write( " f(x) ={e^x if x ≥ 0
\n" ); document.write( " {x + 1 if x < 0
\n" ); document.write( "Determine whether or not that f is continuous at 0.
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Algebra.Com's Answer #781582 by MathLover1(20850)\"\" \"About 
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Suppose \"f%28x%29\" is defined as shown below\r
\n" ); document.write( "\n" ); document.write( "\"f%28x%29+=e%5Ex\" if \"x+%3E=0\"
\n" ); document.write( "........\"x+%2B+1\" if \"x+%3C+0\"\r
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\n" ); document.write( "\n" ); document.write( "Determine whether or not that \"f\" is continuous at \"0\".\r
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\n" ); document.write( "\n" ); document.write( "for the values of \"x+%3E=0\", we have to select the function \"f%28x%29++=+e%5Ex\"\r
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\n" ); document.write( "\n" ); document.write( "right-hand limit: (\"x\" comes from the right, \"x+%3E+a\")\r
\n" ); document.write( "\n" ); document.write( " \"lim%28x-%3E0%2C+f%28x%29+%29=+lim+%28x-%3E0%2Ce%5Ex+%29=1\".......a)\r
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\n" ); document.write( "\n" ); document.write( "for the values of\"+x%3C+0\", we have to select the function\"+f%28x%29++=x+%2B+1\"\r
\n" ); document.write( "\n" ); document.write( "left-hand limit: (\"x\" comes from the left, \"x+%3C+a\")\r
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\n" ); document.write( "\n" ); document.write( " \"lim+%28x-%3E0%2Cx%2B1%29=1\"..........b)\r
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\n" ); document.write( "\n" ); document.write( "\"lim+%28x-%3E0%2Ce%5Ex+%29=+lim+%28x-%3E0%2Cx%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Hence the function is \"continuous\" at \"x+=+0\".\r
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