Question

Question #c747d

Answer 1

It's a two-sided limit, which means that the limit as such does not exist.

Explanation:

`lim_( x->0) 5^x+3^x+2^x-3/x`

The first 3 terms are continuous at `x = 0` so you can just set `x = 0` to get:

`1+ 1+ 1 - lim_( x->0) 3/x`

Now you want to worry about the direction in which you are aproaching zero, because there is a binary change in the sign of x either side of the origin.

So, from the right, where `x>0`:

`lim_( x->0^color(red)(+)) 5^x+3^x+2^x-3/x`

`= 1+ 1+ 1 color(red)(-) lim_( x->0^+) 3/x = - oo`

And from the left, where `x<0`

`lim_( x->0^color(red)(-)) 5^x+3^x+2^x-3/x`

`= 1+ 1+ 1 color(red)(-) lim_( x->0^-) 3/x = + oo`

But, overall, it's a two-sided limit which means that the limit as such does not exist.

Answer 2

`log_e(30)`

Explanation:

I don't like to solve limits using l'Hopital's rule but in this case it is the easiest way to get it.

Making `y = a^x` we have

`x = logy/log a` then `dx = dy/(y log a)` and

`dy/dx = log a y`

so taking the numerator

`d/(dx)(5^x+3^x+2^x-3)=log5 xx 5^x+log3 xx 3^x+log2 xx 2^x`

and the denominator

`d/dx = 1`

then it is equivalent

`lim_(x->0)(5^5+3^x+2^x-3)/x equiv lim_(x->0)(log5 cdot xx 5^x+log3 xx 3^x+log2 xx 2^x)/1`

but `lim_(x->0)a^x = 1` with `a > 0`

so

`lim_(x->0)(5^5+3^x+2^x-3)/x=log(5)+log(3)+log(2)=log_e(30)`