Question

Find a value of a so that the limit exists in a piece wise function?

I'm not sure how to start solving this problem.

For which value a does the lim(fx),x->5 exist where:
`f(x) = { e^(x-a)-2 if x > 5, x^2+5 if x < 5`

Anyone able to tell me where I should start?

I think maybe I set the two piecewise functions equal to each other and evaluate at x=5?

5-ln32 was the answer.

Answer 1

`a=5-ln32`.

Explanation:

Recall that,

`lim_(xto5)f(x)" exists "iff lim_(xto5+)f(x)=lim_(xto5-)f(x)...(star)`.

`"As "x to 5+, x > 5. :. f(x)=e^(x-a)-2.`

`:. lim_(xto5+)f(x)=lim_(xto5+)e^(x-a)-2=e^(5-a)-2..........(star_1)`.

Similarly, `lim_(xto5-)f(x)=lim_(xto5-)x^2+5=5^2+5=30...(star_2)`.

`:." From "(star),(star_1) and (star_2),` we have,

`e^(5-a)-2=30 rArr e^(5-a)=32 rArr lne^(5*a)=ln32`.

`rArr (5-a)=ln32 rArr 5-ln32=a`.

Enjoy Maths.!

Answer 2

see explanation

Explanation:

Yes. The limit exist only when the value of a limit from right equals the value of a limit from left. That means we have to evaluate the value at the point x=5 from each function.

`f_1(x)=e^(x−a)−2;quadquadx>5`
`f_2(x)=x^2+5;quadquadx<5`

`Lim_(xrarr5^+)e^(x-a)-2=e^(5-a)-2`

`Lim_(xrarr5^-)x^2+5=5^2+5=30`

and as I said. These two results must equal:
`e^(5-a)-2=30`
`e^(5-a)=32`
`5-a=ln32`

`5-ln32=a~~1.534`

To check the result I drew a graph. see below.(The closest value of a I could set is 1.5)