Question

Evaluate the limit as x approaches 1 of the quotient of the square root of the quantity x squared plus 24 minus 5 and the quantity x minus 1.

sqr root(x^2 + 24) - 5
Lim as x->1   -----------------------------
                                  x - 1
Can you please explain how I would prove this?

Answer 1

ok, so if you tried to evaluate it, we would get something/0, which is undefined
not helpful

hmm
the next step would be to try and factor something out
well, I don't see what to factor out
the next try would be to use l'hopital's rule

take derivitive of top and bottom
derivitive of top is
`(x)/( √(x^2+24) )`
derivitive of bottom is 1
so we get
`((x)/( √(x^2+24) ))/(1) =(x)/( √(x^2+24) )`
if we input 1 for x we get


`(1)/( √(1^2+24) )`=
`(1)/( √(1+24) )`=
`(1)/( √(25) )`=
`(1)/(5)`
we know it can't be
`(-1)/(5)` because if we evaluated it at x=0 and x=2, we see that it is above 0


the as x->1,
`( √(x^2+24) -5)/(x-1)`=1/5

Answer 2

sqr root(x^2 + 24) - 5
Lim as x->1 -----------------------------
x - 1
when you will put x=1 in denominator that time it will become 0.
that whole part will become infinity
lim x--->1
`\sqrt{ (x^(2) +24)-5}`/x-1

In numerator put x=1 u will get
lim x--->1 √(1)²+24/1-1
lim x--->1 √25/0

so that the resulting fraction approaches ∞