1) You need to substitute -3 for x in equation under limit such that:

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = (5-sqrt(22+3))/(1-sqrt(-3+4))`

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = (5-5)/(1-1) = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = lim_(x-gt-3) ((5-sqrt(22-x))')/((1-sqrt(x+4))') `

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = lim_(x-gt-3) (1/(2sqrt(22-x)))/(-1/(2sqrt(x+4)))`

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = lim_(x-gt-3) (-sqrt(x+4))/(sqrt(22-x))`

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1) You need to substitute -3 for x in equation under limit such that:

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = (5-sqrt(22+3))/(1-sqrt(-3+4))`

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = (5-5)/(1-1) = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = lim_(x-gt-3) ((5-sqrt(22-x))')/((1-sqrt(x+4))') `

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = lim_(x-gt-3) (1/(2sqrt(22-x)))/(-1/(2sqrt(x+4)))`

`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = lim_(x-gt-3) (-sqrt(x+4))/(sqrt(22-x))`

You need to substitute -3 for x in equation under limit such that:

`lim_(x-gt-3) -(sqrt(x+4))/(sqrt(22-x)) =- (sqrt(-3+4))/(sqrt(22+3))`

`lim_(x-gt-3) (-sqrt(x+4))/(sqrt(22-x)) = -1/5`

**Hence, evaluating the limit yields`lim_(x-gt-3) (5-sqrt(22-x))/(1-sqrt(x+4)) = -1/5.` **

2) You need to substitute 1 for x in equation under limit such that:

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = (2-sqrt(5-1))/(3-sqrt(8+1))`

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = (2-sqrt(4))/(3-sqrt(9))`

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = (2-2)/(3-3) = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = lim_(x-gt1) ((2-sqrt(5-x))')/((3-sqrt(8+x))') `

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = lim_(x-gt1)(1/(2sqrt(5-x)))/(-1/(2sqrt(8+x)))`

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = lim_(x-gt1) (-sqrt(8+x))/(sqrt(5-x))`

You need to substitute 1 for x in equation under limit such that:

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = (-sqrt(8+1))/(sqrt(5-1))`

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = (-sqrt(9))/(sqrt(4))`

`lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = -3/2`

**Hence, evaluating the limit yields `lim_(x-gt1) (2-sqrt(5-x))/(3-sqrt(8+x)) = -3/2.` **

3) You need to substitute 4 for x in equation under limit such that:

`lim_(x-gt4)(1-sqrt(x-3))/(2-sqrt x) = (1-sqrt(4-3))/(2-sqrt 4)`

`lim_(x-gt4) (1-sqrt(x-3))/(2-sqrt x) = (1-sqrt(1))/(2-2)`

`lim_(x-gt4) (1-sqrt(x-3))/(2-sqrt x) = (1-1)/(0) = 0/0`

Since the limit is indeterminate, you may use l'Hospital's theorem such that:

`lim_(x-gt4) (1-sqrt(x-3))/(2-sqrt x) =lim_(x-gt4) ((1-sqrt(x-3))')/((2-sqrt x)')`

`lim_(x-gt4) (1-sqrt(x-3))/(2-sqrt x) = lim_(x-gt4) (-1/(2sqrt(x-3)))/(-1/(2sqrt(x)))`

`lim_(x-gt4) (1-sqrt(x-3))/(2-sqrt x) = lim_(x-gt4) (-sqrt(x))/(-sqrt(x-3))`

You need to substitute 4 for x in equation under limit such that:

`lim_(x-gt4) (1-sqrt(x-3))/(2-sqrt x) = lim_(x-gt4) (-sqrt(4))/(-sqrt(4-3)) = 2/1`

**Hence, evaluating the limit yields `lim_(x-gt4) (1-sqrt(x-3))/(2-sqrt x) = 2.` **