Question

Evaluate the indicated limit, if it exists. If it does not exist, explain why it doesn’t. Assume that lim x-> 0 sin x/x = 1

Answer 1

it is a well known limit
\[\lim_{x\to 0}\frac{\sin(x)}{x}=1\]

Answer 2

but the proof is not obvious
look in any intro calculus book, it will be there using a geometric argument and the "squeeze theorem"

Answer 3

1) lim x-> 2 (x-5/x^2 + 4)
2) lim x-> 3 (x^2 - x - 6/x - 3)
3) lim h-> 0 ((2 + h)^2 - 4/h)

Answer 4

1) replace \(x\) by 2

Answer 5

ok

Answer 6

2) factor as
\[\frac{(x-3)(x+2)}{x-3}=x+2\] then replace \(x\) by 3

Answer 7

okay

Answer 8

3) expand get
\[\frac{(2+h)^2-4}{h}=\frac{4+4h+h^2-4}{h}=\frac{4h+h^2}{h}=4+h\] then replace \(h\) by 0

Answer 9

ok

Answer 10

for number 4) its this one lim t-> -2 (1/2 + 1/t over 2 + t)

Answer 11

and 5) lim x->4 + sqrt(16 - x^2)

Answer 12

\[\frac{1}{2}+\frac{1}{t}=\frac{t+2}{2t}\] divide by \(t+2\) and get
\[\frac{1}{2t}\] replace \(t\) by \(-2\)

Answer 13

ok

Answer 14

\[\lim_{x\to 4^+}\sqrt{16-x^2}\] does not exist because if \(x>4\) then \(16-x^2<0\) and you cannot take the square root of a negative number

Answer 15

okay thanks alot :)