lim x tend to a then log(1+3f(x))/2f(x) is????

Question

when f(x) is continous function such that its graph passes through the point (a,0).

Answer 1

Note: I'm assuming you mean the natural logarithm (base e), not the common logarithm (base 10). Otherwise, the approach is the same, but the answer will be different.

Since f(x) is continuous,

lim x->a f(x) = f(a) = 0

because (a,0) is on the graph of f. That means the y-value 0 corresponds to the x-value a, so f(a)=0.

To get the limit of log(1+3f(x)) / (2f(x)), plug in 0 for f(x), since that is its value as x -> a. You get

log(1+0) / (2*0) = 0/0

so you must use l'Hospital's Rule. We need to assume that f is differentiable for this to be valid.

Assuming it is, we differentiate the numerator and denominator, and get

3f'(x)/[1+3f(x)] / [2f'(x)]

which simplifies to

3 / [2(1+3f(x))].

To compute the limit of this, as x->a, plug in f(a)=0 for f(x).

Answer 2

Is the logarithm here common log (base 10) or natural log (base e)?

Since lim ln (y+1)/y = 1 if y --> 0,
lim ln (3f + 1)/(3f) = 1
and lim (3f + 1)/(2f) = 3/2 when f --> 0, which is the case here

In the case of common log, realize that log x = ln x / ln 10,
so the limit is 3 / (2 ln 10).