<p>If f(x)=x^3-x&2-3x-1 and h(x)=f(x)/g(x), evaluate limit as x->0 for (3h(x) + f(x) - 2g(x))</p>
<p>assuming you know h(x):
h is continuous everywhere except x=-1
limit x-> infinity, h(x) = infinity
lim x-> -1, h(x) = 1/2</p>
<p>please please please help</p>
<p>I’m not too entirely sure but I’ll give it a try: </p>
<p>Using the 2nd assumption of h, it looks like g(x) gets smaller and smaller as x increases. So it gets bigger and bigger as x decreases (towards 0). So with that, 3(h(x)) goes to 0, and -2(g(x) goes to 0, so you’re left with f(x). In this case, the answer is -1. </p>
<p>NOT sure, but this is what I think.</p>