1.) lim x --> 1(left) (x^2 + |x| -2)/(|1-x|)(adsbygoogle = window.adsbygoogle || []).push({});

Since the limit is from the left, i made all absolute values pnegative, therefore numerator = x^2 - x - 2, and similarly denom. = (x-1). Then, by inspection, the limit would equal to + infinite.

2.) Find numbers a and c such that lim x --> 0 x/(sqrt(ax + c) -3) = 2

Multiplying first by the conjugate, and then assuming a value for either a or c (in my case, i let c = 9). Then, solving for a, i got 3. Is this assumption allowed/necessary?

3.) lim (x-->0) f(x^2)/x^2, where f(0)=0 and f'(0)=3.

No idea.

4.) Evaluate f'(P), where f(P) = tan(3P + sinP))

No idea.

5.) Prove: If lim x--> 6 f(x)g(x) exists, then limit must be f(6)g(6).

6.) Prove: If f is continuous at 5 and f(5)=2 and f(4)=3, then lim x-->2f(4x^2 - 11) =2

Thank you for absolutely any input. These are only a selected few (out of very many) that i was unsuccessful at solving.

Thank you again.

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# Homework Help: Calculus I Questions

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