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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 XXXX,</p><p>lim(x-&XX;a) f(x) =(a^2 -5a+4)/(a^2 -1)</p><p><strong XXXXX="redactor-inline-XXXXXXXXX">2.</strong> The XXXXX is attached.</p><p><em>Explanation</em>:</p><p> x=0 : f(x) =(X-4)/(X+1) = -4/X = -X</p><p>x=X : f(x)=(X-X)/(4+1) = 0/X = 0</p><p>x>4 => x=(4+a) XXX any XXXXXXXX XXXXX of a :f(x)=(4+a-X)/(X+a+1) = a/(a+X)<1</p><p>x=-X :f(x)=(-1-X)/(-1+1) = -5/0 = -(XXXXXXXX)</p><p>x<-X => x=(-X-b)XXX any positive value ofb :f(x)=(-1-b-X)/(-X-b+1) = -(b+5)/-b = X+ 5/b &XX; 1</p><p>XXXX b =infinity => f(x) = X + 1/(XXXXXXXX) = 1+0 = X</p><p><strong XXXXX="redactor-inline-converted">X.</strong> XXX(x-&XX;-X) f(x) =XXX(x->-X)(x-4)/(x+X) =(-1-4)/(-X+X) = -X/X = -(infinity)</p><p><strong class="XXXXXXXX-inline-converted">X.</XXXXXX>f(x) =(x^2 -5x+X)/(x^X -X)</p><p>(x^2 -XX+4) XX XXXXXXXXXX at x=-X and also(x^2 -1) is XXXXXXXXXX XX x=-4<XX></p><p>Againlim(x-&XX;-4) f(x) =XXX(x-&XX;-X) (x^X -XX+4)/(x^X -X)</p><p>=XXX(x-&XX;-X)(x-X)/(x+X) = (-X-X)/ (-4+1) = -X/-X = X/3</p><p>Also f(-X) =[(-4)^X -5(-4)+4]/[(-4)^2 -1] = XX/15 = X/X</p><p>lim(x->-4) f(x) =f(-X)<br></p><p>XXXX <strong>f(x) XXXXXXXXXXXX XX x=-4</strong></p><p><br></p>">
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