1. ## limits plz help

1.Find the value of Lim x->0+ (cos x-1)/x^2
(a)1/4 (b)-2 (c) infinity (d) -1/2 (e) none

2.find the value of limit lim x->inf x sin(1/x)
(a)0 (b)1 (c)2 (d)4 (e)none

3.find the value of limit lim x>infinity x^(2/x)
(a)2 (b) e (c) ln 2(d)1 (e) none

4.find the value of lim x->inf ( ln x)^2/x
(a) - inf (b) -1 (c) 0 (d) e (e) none

5.find the value of limit lim x->1 (1/(x-1) - 1/ln x)
(a)-1 (b)0 (c) 1/2 (d) -1/2 (e) none

6.find the value of limit lim x->pi - (sin x/ 1-cos x)
(a)1 (b)-1 (c) infinity (d) -infinity (e) none

2. Originally Posted by bobby77
1.Find the value of Lim x->0+ (cos x-1)/x^2
(a)1/4 (b)-2 (c) infinity (d) -1/2 (e) none
This helps to use the "extension of the real line" for this problem.
You have,
$\lim_{x\to\infty}\frac{\cos x-1}{x}\cdot \frac{1}{x}$
The first product exists (it is a standard limit with value of 1). The second does not exist but increases without bound. Thus, the answer is (c)

2.find the value of limit lim x->inf x sin(1/x)
(a)0 (b)1 (c)2 (d)4 (e)none
Let, $g=1/x$, $f=(1/x)\sin x$ then, the limit of,
$\lim_{x\to\infty}f\circ g$ can be found by evaluating the inner limit, $\lim_{x\to\infty}g=0$.So the limit of,
$\lim_{x\to\infty}f\circ g=\lim_{x\to 0}\frac{\sin x}{x}$ by limit composition rule.

Originally Posted by bobby77

3.find the value of limit lim x>infinity x^(2/x)
(a)2 (b) e (c) ln 2(d)1 (e) none
You can express,
$x^{2/x}=\exp (\ln x^{2/x})=\exp \left( \frac{2\ln x}{x} \right)$
The limit of the inner function exists and is zero because any polynomial overtakes a logarithm. Then $e^0=1$ by the limit composition rule.