onegirl 2 years ago 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

1. satellite73

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

2. satellite73

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

3. onegirl

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)

4. satellite73

1) replace $$x$$ by 2

5. onegirl

ok

6. satellite73

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

7. onegirl

okay

8. satellite73

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

9. onegirl

ok

10. onegirl

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

11. onegirl

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

12. satellite73

$\frac{1}{2}+\frac{1}{t}=\frac{t+2}{2t}$ divide by $$t+2$$ and get $\frac{1}{2t}$ replace $$t$$ by $$-2$$

13. onegirl

ok

14. satellite73

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

15. onegirl

okay thanks alot :)