Question

How do you find the limit `(1/t+1/sqrtt)(sqrt(t+1)-1)` as `t->0^+`?

Answer 1

`lim_(t rarr 0^+)( 1/t + 1/sqrt(t) )( sqrt(t+1) - 1 ) = 1/2`

Explanation:

Let `L = lim_(t rarr 0^+)( 1/t + 1/sqrt(t) )( sqrt(t+1) - 1 )`

`:. L = lim_(t rarr 0^+) ( 1/t + 1/t^(1/2) )( (t+1)^(1/2) - 1 )`
`:. L = lim_(t rarr 0^+) ( 1/t + t^(1/2)/t )( (t+1)^(1/2) - 1 )`
`:. L = lim_(t rarr 0^+) ( ( 1 + t^(1/2) )( (t+1)^(1/2) - 1 ) ) / t`

This is now in the indeterminate form `0/0`, so we can apply L'Hospital's Rule:

`:. L = lim_(t rarr 0^+) ( ( 1 + t^(1/2) ) d/dt( (t+1)^(1/2) - 1 ) + d/dt( 1 + t^(1/2) ) ( (t+1)^(1/2) - 1 ) ) /1`
`:. L = lim_(t rarr 0^+) {( 1 + t^(1/2) ) ( 1/2(t+1)^(-1/2) ) + ( 1/2t^(-1/2) ) ( (t+1)^(1/2) - 1 ) )}`
`:. L = lim_(t rarr 0^+) { ( 1 + sqrt(t) )/(2sqrt(t+1)) + ( sqrt( t+1) - 1 ) /(2sqrt(t) ) }`

Let `L_1 = lim_(t rarr 0^+) ( 1 + sqrt(t) )/(2sqrt(t+1))`, and, `L_2= lim_(t rarr 0^+)( sqrt( t+1) - 1 ) /(2sqrt(t) )`

Then `L = L_1 + L_2`

We can evaluate `L_1` directly:

`:. L_1 =( 1+0) /(2sqrt(0+1)) =1/2`

And I won't do the maths because it it is really tedious, but a couple more application of L'Hospital's Rule will show that

`L_2 = 0`

Hence, `L = lim_(t rarr 0^+)( 1/t + 1/sqrt(t) )( sqrt(t+1) - 1 ) = 1/2`

We can verify this by plotting a graph of the function

`f(x) = ( 1/x + 1/sqrt(x) )( sqrt(x+1) - 1 )`

graph{( 1/x + 1/sqrt(x) )( sqrt(x+1) - 1 ) [-0.811, 1.3165, -0.191, 0.873]}