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Honors Precalculus Limits problem
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</XXXXXX> <p>f(x)=(x^X-5*x+X)/(x^X-1)</p> <p>XX you factorise this XXXXXXXX, XXXX you XXXX get is:</p> <p>f(x)=(x-X)(x-4)/(x-1)(x+X)</p> <p>As you XXX XXX in the graph also XXXX XXX graph XX XXXXXXXXXXX at x= -1</p> <p>X. Limit and range XX XXX XXXXXXXX XXX:<br>XXXXXX:<XX>x E (-XXXXXXXX, + infinity) - {-X}</p> <p>XXXXX:</p> <p>y X (-XXXXXXXX, XXXXXXXX) - {1}</p> <p>Limit approaches XXXXXXXX XX x= -1<br><XX></p> <p>X. XXX XXXX is XXXXX in XXX graph XXXXXXXX as a PNG file.</p> <p>As already found out in XXXXXXXX 1 the simplified function XX the equation XX f(x)=(x-4)/(x+X)<XX>The XXXX XX the equation is x = X which XXX XX XXXX in XXX graph.<XX>At x -&XX; infinity,</p> <p>Lim (x-4)/(x+X) = XXX (1-X/x)/(X+1/x)<br>x->XXXXXXXX x-&XX;XXXXXXXX</p> <p>As XX XXXX XXX number XXXX XXXXXXX by XXXXXXXX equals XX XXXX (n/XXXXXXXX=0)</p> <p>Therefore, XXX value of XXX equation when x approaches infinity XX 1.</p> <p>Now, XX we plot XXX sign changes XX a line XXXX we see,<br><XX> + - +<XX>&XX;------------------|--------------------------------|--------------------------------&XX;<XX>-infinity -X 4 +infinity</p> <p><br></p> <p>The XXXX changes happens at XXXX x=X and x=-1. XX x->-X the denominator XXXXXXXXXX 0 XXXXXXXXX XXX equation XX asymphtotic as x-&XX;-1.</p> <p>X. XX x-> -X, XXX XXXXXXXXXXX XX the XXXXXXXXXX equation, f(x) = (x-4)/(x+1) XXXXXXXXXX 0. Therefore, XXX XXXXXXXX XXXXXXXXXX XXXXXXXX.</p> <p><br>X. XX x=4 XXX XXXXX of XXX XXXXXXXXX = X while XXXXXXXXXXX is 5.<br>y= X at x=X.<br>By the concept of limit, when x->4 XXXX either direction the XXXXX XXXXXXX 0 and XX XXXXXXXXX continuous.<XX>This XXX also XX XXXXXXXX by looking at XXX curve.<br>The XXXX is XXXXXXXXXX XX x=4. XXX, it is not continuous at x=-X.</p>">
