Answer to Question #201496 in Calculus for Sanjeshni

Question

1. Use the Squeezing Theorem to show that

lim

X approaches to +sum to infinity

Sin x divide by x =0

2. Use the result in part 1. to find

lim

X approaches to +sum to infinity

Cos πx divide by square root of xsquared +x sinx

Answer 1

Solution:

(1):

We know that $\sin x \in [-1,1]$

even when x tends to infinity the value of sin x lies between -1 and 1., which is defined value

when x tends to infinity, 1/x will tend to 0.

So, when x tends to infinity $\frac{\sin x}{x}$ will be some finite value [lying between -1 and 1)*(0)]

Thus, by squeeze theorem,

$\lim_{x\rightarrow \infty}\frac{\sin x}{x}=0$

(2):

$\lim _{x\to \infty \:}\left(\dfrac{cos\left(\pi \:x\right)}{\sqrt{x^2+x\:sin\:x}}\right)$

$=\lim _{x\to \infty \:}\left(\dfrac{\dfrac{\cos \left(\pi x\right)}{x}}{\sqrt{1+\dfrac{\sin \left(x\right)}{x}}}\right)$

$=\dfrac{\lim _{x\to \infty \:}\left(\dfrac{\cos \left(\pi x\right)}{x}\right)}{\lim _{x\to \infty \:}\left(\sqrt{1+\frac{\sin \left(x\right)}{x}}\right)}$

$=\dfrac0{\sqrt{1+0}}$ [Numerator becomes 0 after applying limit and using squeeze theorem, while we use part 1 in denominator]

$\\=0$