Question

Calculate the following limit `lim_(x->0) (a^x - b^x)/x`? The answer should be `log(b/a)`.

Answer 1

`lim_(x->0) (a^x-b^x)/x=ln(a/b)=log(a/b)`

Explanation:

We want to find `lim_(x->0) (a^x-b^x)/x`.

Plugging in the limiting value, we get

`(a^0-b^0)/0=(1-1)/0=0/0`

This is an indeterminate form, so we can use l'Hopital's rule

`lim_(x->0)(a^x-b^x)/x=lim_(x->0) (d/dx(a^x)-d/dx(b^x))/(d/dxx)=lim_(x->0)(a^xlna-b^xlnb)=a^0lna-b^0lnb=lna-lnb=ln(a/b)-=log(a/b)`

Answer 2

See below.

Explanation:

`(a^x-b^x)/x = (a^x-1)/x-(b^x-1)/x = (a^(0+x)-a^0)/x-(b^(0+x)-b^0)/x`

now

`lim_(x->0)(a^x-b^x)/x =lim_(x->0) (a^(0+x)-a^0)/x-lim_(x->0)(b^(0+x)-b^0)/x = lna-lnb = ln(a/b)`

Answer 3

`lim_(x rarr 0) (a^x-b^x)/x = ln(a/b)`

Explanation:

We seek the limit:

`L = lim_(x rarr 0) (a^x-b^x)/x`
`\ \ = lim_(x rarr 0) (a^x-1-b^x-(-1))/x`
`\ \ = lim_(x rarr 0) {(a^x-1)/x -(b^x-1)/x }`
`\ \ = L_a - L_b \ \`, say ..... [A]

Where:

`L_alpha = lim_(x rarr 0) (alpha^x-1)/x`

And, we now have two different ways of calculating this limit:

Method:1

Using the limit definition of the derivative we have:

`f'(a) = lim_(h rarr 0) (f(x)-f(a))/(x-a)`

We note that:

`L_alpha = lim_(x rarr 0) (alpha^x-1)/x`
`\ \ \ \ = lim_(x rarr 0) (alpha^x-alpha^0)/(x-0)`
`\ \ \ \ = lim_(x rarr 0) (f(x)-f(0))/(x-0) \ \ \ \` where `f(x)=alpha^x`
`\ \ \ \ = f'(0)`

And if `f(x)=alpha^x` then we have:

`f(x) =e^ln(alpha)x`
`:. f'(x)= lnalpha \ e^ln(alpha)x = ln alpha \ alpha^x`
`=> f'(0) = ln alpha \ alpha^0 = ln alpha`

Method:2

Let `y = alpha^x-1 => 1+y =alpha^x`, then taking logarithms we have:

`ln(1+y) = ln alpha^x`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \ = xln alpha`
`:. x = (ln(1+y))/(ln alpha)`

So we can write:

`L_alpha = lim_(x rarr 0) (alpha^x-1)/x`
`\ \ \ \ = lim_(y rarr 0) (y)/((ln(1+y))/(ln alpha)) \ \ \ \ because y rarr 0` as `x rarr 0`
`\ \ \ \ = lim_(y rarr 0) (lnalpha)/(1/yln(1+y))`
`\ \ \ \ = lim_(y rarr 0) (lnalpha)/(ln(1+y)^(1/y))`
`\ \ \ \ = (lnalpha)/(lim_(y rarr 0) {ln(1+y)^(1/y)}) \ \ \ \` properties of limits
`\ \ \ \ = (lnalpha)/( ln {lim_(y rarr 0)(1+y)^(1/y)}) \ \ \ \` monotonicity of logarithms
`\ \ \ \ = (lnalpha)/( ln e) \ \ \ \ because lim_(y rarr 0)(1+y)^(1/y)=e`
`\ \ \ \ = ln alpha \ \ \ \ because lne=1`

Consistent with the previous method:

Then using this result in conjunction with [A], we have:

`L = lna - ln b`
`\ \ = ln(a/b) \ \ \ \` properties of logarithms