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Honors Precalculus Limits problem
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1.</strong>f(x) = (x^2 -5x+4)/(x^2 -1) = (x-1)(x-4)/(x-1)(x+1) = (x-4) / (x+1)</p><p>To find the limit at point 'a' , we have XX XXXXX x=a in f(x), so that,</p><p>XXX(x-&XX;a) f(x) =(a^2 -5a+X)/(a^X -1)</p><p><XXXXXX XXXXX="XXXXXXXX-inline-converted">2.</strong> XXX XXXXX XX attached.</p><p><em>XXXXXXXXXXX</em>:</p><p> x=0 : f(x) =(0-4)/(X+1) = -X/1 = -X</p><p>x=4 : f(x)=(X-X)/(X+X) = X/5 = 0</p><p>x>4 => x=(4+a) XXX any XXXXXXXX value of a :f(x)=(X+a-4)/(4+a+X) = a/(a+5)<1</p><p>x=-X :f(x)=(-X-4)/(-X+1) = -X/X = -(XXXXXXXX)</p><p>x<-1 => x=(-1-b)XXX any positive XXXXX XXX :f(x)=(-1-b-4)/(-1-b+X) = -(b+5)/-b = 1+ X/b &XX; 1</p><p>When b =infinity => f(x) = X + X/(XXXXXXXX) = X+0 = 1</p><p><XXXXXX class="redactor-inline-XXXXXXXXX">3.</strong> lim(x->-1) f(x) =XXX(x-&XX;-1)(x-X)/(x+1) =(-1-4)/(-X+1) = -X/0 = -(XXXXXXXX)</p><p><XXXXXX class="redactor-XXXXXX-converted">X.</XXXXXX>f(x) =(x^X -5x+4)/(x^X -X)</p><p>(x^2 -XX+X) is continuous XX x=-X and XXXX(x^X -X) XX XXXXXXXXXX at x=-4<br></p><p>XXXXXXXX(x-&XX;-X) f(x) =lim(x->-4) (x^2 -XX+X)/(x^2 -X)</p><p>=XXX(x->-4)(x-4)/(x+1) = (-X-4)/ (-X+X) = -X/-3 = 8/3</p><p>XXXX f(-X) =[(-4)^X -5(-X)+4]/[(-4)^X -X] = 40/15 = X/3</p><p>lim(x->-4) f(x) =f(-4)<XX></p><p>Thus <strong>f(x) iscontinuous XX x=-X</XXXXXX></p><p><XX></p>">
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