How can I find numbers a and b such that the limit as x approaches 0 is 1.?

This is the given equation:


the lim as X approaches 0 ( the sqrt of(aX+b) ) minus 2 divided by X is equal to 1














This is a homework assignment I need help with this one. I can do the others well. But on this one I don't know weather to check for continuity or not. I need help with an approach to solving this limit for numbers a and b. Two X's are in this equation one in the numerator (after the a) and the only one in the denominator. Any help is appreciated. Please let me know immediately if you can not understand anything concerning this problem. Thanks.How can I find numbers a and b such that the limit as x approaches 0 is 1.?
Multiply by x on both sides to get it out of the denominator.





sqrt(ax+b) - 2 = x


sqrt(ax+b) = x+2


ax+b = (x+2)^2 = x^2 +4x+4


Now, substitute in x = 0.


b = 4.





You'll see that if you put 4 back in, you have 0/0. But it's a 0/0 that approaches 1.





So your answer is 4.How can I find numbers a and b such that the limit as x approaches 0 is 1.?
i have already forgotten this lesson but as i can remember, you should first check for the continuity. then try to equate the left-hand and right-hand limits.=)
Ok, I'll take a shot at it, but first I need to know, is everything divided by x, or just the 2 divided by x?


That is, do you have


(sqrt (ax + b) - 2)/x or


sqrt (ax + b) - (2/x)?


If the former, I'd begin by rationalizing the numerator. If the latter, I don't think there would be a solution.
neat problem...





normally when you try to evaluate a limit you start out by substituting..





If in this case you substitute and get k/0, then the limit will not exist (it will go to +inf from one side, and -inf from the other.) therefore numerator needs to be 0 as well. When this happens we'll have a common factor of x in numerator and denominator that will cancel when you simplify which will allow to evaluate the limit. With that in mind.





sqrt(ax+b) - 2 = 0


sqrt(ax+b) = 2


ax+b = 4





Substituting 0 + 4 = b b=4 using that





lim x-%26gt;0 (sqrt(ax+4)-2))/x





multiply numerator and denominator by the conjugate sqrt(ax+4)+2





lim x-%26gt;0 (sqrt(ax+4)-2)(sqrt(ax+4)+2)/(x(sqrt(ax+鈥?br>

lim x-%26gt;0 (ax+4-4)/(x(sqrt(ax+b)+2))


lim x-%26gt;0 ax/(x(sqrt(ax+4)+2))


lim x-%26gt;0 a/(sqrt(ax+4)+2)





evaluate the limit





a/(sqrt(0+4)+2)


a/4





so a/4 must be 1





a = 4





lim x-%26gt;0 (sqrt(4x+4)-2)/x = 1





Should be noted that if b is not 4 the limit is undefined. if be is four then the limit would be a/4
rationalize the numerator: multiply both numerator and denominator by [sqrt(aX+b) + 2]. Numerator will then be [(aX+b) - 4] and denominator will be X*[sqrt(aX+b) + 2]. Take b = 4, then cancel an X in the numerator and denominator. Let X go to 0. Numerator goes to (is) a . Denominator goes to [sqrt(4) + 2] = 4. To get the limit to be 1, take a = 4.





Enjoy.