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need to find: lim as x > 0 of 4(e^2x  1) / (e^x 1) Try splitting the limit for the numerator and denominator lim lim x>0 4(e^2x1) (4)x>0 (e^2x1) ______________ = ________________ lim lim x>0 e^X1 x>0 e^x1 
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Find the following limits algebraically or explain why they don’t exist. lim x>0 sin5x/2x lim x>0 1cosx/x lim x>7 x7/x7 lim x>7 (/x+2)3/x7 lim h>0 (2+h)^38/h lim t>0 1/t  1/t^2+t 
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Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? and lim x>0 = + infinity? lim x>1 (x^2  5x + 6)/(x^2  3x + 2) I get 2/0, so lim x> 1+ =  infinity? and lim 
calc
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? and lim x>0 = + infinity? lim x>1 (x^2  5x + 6)/(x^2  3x + 2) I get 2/0, so lim x> 1+ =  infinity? and lim 
Calc Please Help
Are these correct? lim x>0 (x)/(sqrt(x^2+4)  2) I get 4/0= +/ infinity so lim x>0+ = + infinity? and lim x>0 = + infinity? lim x>1 (x^2  5x + 6)/(x^2  3x + 2) I get 2/0, so lim x> 1+ =  infinity? and lim 
Calculus
Find the limit. lim 5x/(x^225) x>5 Here is the work I have so far: lim 5x/(x^225) = lim 5x/(x5)(x+5) x>5 x>5 lim (1/x+5) = lim 1/10 x>5 x>5 I just wanted to double check with someone and see if the 
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i need some serious help with limits in precalc. here are a few questions that i really do not understand. 1. Evaluate: lim (3x^32x^2+5) x> 1 2. Evaluate: lim [ln(4x+1) x>2 3. Evaluate: lim[cos(pi x/3)] x>2 4. 
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How do we evaluate limit of, lim x> 0 [ln(x+1)/( (2^x)  1)] I tried using the substitution x+1 = e^k , when x tends to 0 so does k, which gave out, lim k>0 [ k/((2^((e^k)  1)) 1 ) ] which I simplified into( for the ease 
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Evaluat d 4lowing1.lim x/x x>0 2.lim x>1 sqrt(x^2+2 sqrt3)/x1 3.lim n>~ f(n)=(1+1/n)^sqrtn 4. limx>0 f(x)= (12^x3^x4^x+1)/xtanx 5. Lim x>3 (x^n3^n)^n/(n3)^n 
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If y = 3 is a horizontal asymptote of a rational function, which must be true? lim x→ 3 f(x) = 0 <<my answer lim x→ 0 f(x) = 3 lim x→ ∞ f(x) = 3 lim x→ 3 f(x) = ∞