Author Topic: how to do this question? (Read 824 times) Tweet Share

lkkdfjflkdsjlk · « **on:** March 12, 2011, 11:09:15 pm »

using limit theorems and standard results,find the limit, if it exists

lim x approaches infinity cos(1+x^2) divided by x^2

Mao · « **Reply #1 on:** March 13, 2011, 12:00:10 am »

0.

Cos is between -1 and 1. x^2 approaches infinity as x approaches infinity.

lkkdfjflkdsjlk · « **Reply #2 on:** March 13, 2011, 12:18:25 am »

Quote from: Mao on March 13, 2011, 12:00:10 am

0.

Cos is between -1 and 1. x^2 approaches infinity as x approaches infinity.

thanks mao....

but how would i show the working out?

/0 · « **Reply #3 on:** March 13, 2011, 04:46:41 am »

Formal way is by the "squeeze theorem", which says that if $f(x) \leq g(x) \leq h(x)$ for all $x$ and $\lim_{x \to \infty} f(x)=\lim_{x\to \infty} h(x)=c$ , then since $g(x)$ is sandwiched between the other two, it must have the same limit: $\lim_{x\to \infty}g(x)=c$ .

Here, since $-1 \leq \cos{(1+x^2)}\leq 1$ , we have

$-\frac{1}{x^2} \leq \frac{\cos{(1+x^2)}}{x^2}\leq \frac{1}{x^2}$

We know that $\pm \frac{1}{x^2} \to 0$ as $x \to \infty$ . Since $\frac{\cos{(1+x^2)}}{x^2}$ is sandwiched between these two terms, $\frac{\cos{(1+x^2)}}{x^2} \to 0$ .

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Author Topic: how to do this question? (Read 824 times) Tweet Share

lkkdfjflkdsjlk

how to do this question?

Mao

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lkkdfjflkdsjlk

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