Question

What is the limit of `(2^x -32)/(x-5 )` as x approaches `5`?

Answer 1

`32 ln(2) ~~ 22.1807`

Explanation:

Use L'Hôpital's rule:

If functions `f` and `g` are differentiable on an open interval `I` except possibly at `c in I` and all of the following hold:

• `lim_(x->c) f(x) = lim_(x->c) g(x) = 0` or `+-oo`
• `g'(x) != 0` for all `x in I "\" { c }`
• `lim_(x->c) (f'(x))/(g'(x))` exists

Then:

`lim_(x->c) f(x)/g(x) = lim_(x->c) (f'(x))/(g'(x))`

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In our example, let `f(x) = 2^x-32` and `g(x) = x-5`.

Note that:

`f'(x) = d/(dx) (2^x-32) = d/(dx) ((e^(ln(2)))^x - 32)`

`=d/(dx) (e^(ln(2)*x)-32) = ln(2) e^(ln(2)*x) = ln(2)*2^x`

`g'(x) = d/(dx) (x-5) = 1`

These are both differentiable in the whole of `RR`, let alone any open interval we might choose containing `c=5`.

• `lim_(x->5) f(x) = lim_(x->5) (2^x-32) = 32 - 32 = 0`
• `lim_(x->5) g(x) = lim_(x->5) (x-5) = 5 - 5 = 0`
• `g'(x) = 1 != 0` for all `x in RR`
• `lim_(x->5) (f'(x))/(g'(x)) = (ln(2)*2^5)/1 = 32 ln(2)`

Hence:

`lim_(x-5) (f(x))/(g(x)) = lim_(x->5) (f'(x))/(g'(x)) = 32 ln(2) ~~ 22.1807`

graph{((2^x-32)/(x-5)-y)((x-5)^2+(y-32ln(2))^2-0.1)=0 [-32.34, 32.6, -3.77, 28.7]}