how to find the following limit :

lim lnx/lgx
x->infinity

If lg(x) = log2x, then since

lgx = lnx/ln2,

lnx/lgx = ln2 for all x.

To find the limit of lnx/lgx as x approaches infinity, we can use L'Hôpital's Rule. L'Hôpital's Rule states that if the limit of the ratio of two functions exists but gives an indeterminate form (such as infinity divided by infinity), then taking the derivative of the numerator and denominator separately may help to evaluate the limit.

Starting with the given limit:
lim (x->∞) (lnx/lgx)

We can rewrite the expression using the logarithm properties:
lim (x->∞) ln(x) / ln(g(x))

Now, we can apply L'Hôpital's Rule by taking the derivative of the numerator and denominator individually. The derivative of ln(x) is 1/x, and the derivative of ln(g(x)) is g'(x)/g(x):

lim (x->∞) (1/x) / (g'(x)/g(x))

Simplifying further, we multiply by the reciprocal of the denominator:
lim (x->∞) (g(x)/g'(x)) * (x/1)

Now, if the denominator g'(x) approaches infinity as x goes to infinity, and g(x) is a non-zero constant, the limit becomes:
lim (x->∞) (g(x)/g'(x)) * (x/1) = ∞ (infinity)

However, if the denominator g'(x) approaches zero as x goes to infinity, and g(x) is a non-zero constant, the limit becomes:
lim (x->∞) (g(x)/g'(x)) * (x/1) = 0

Therefore, to determine the limit of lnx/lgx as x approaches infinity, we need to know the function g(x) and its derivative g'(x). Without specifying the function g(x), it is not possible to find a definite answer to the limit.