# Solving Limit Problems To do this we need to be quite clever, and to employ some indirect reasoning.The indirect reasoning is embodied in a theorem, frequently called the squeeze theorem.To do the hard limit that we want, $\lim_ (\sin x)/x$, we will find two simpler functions $g$ and $h$ so that $g(x)\le (\sin x)/x\le h(x)$, and so that $\lim_g(x)=\lim_h(x)$.

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This theorem can be proved using the official definition of limit.

We won't prove it here, but point out that it is easy to understand and believe graphically.

It is not so easy to see directly, that is algebraically, that $\lim_x^2\sin(\pi/x)=0$, because the $\pi/x$ prevents us from simply plugging in $x=0$.

The squeeze theorem makes this "hard limit'' as easy as the trivial limits involving $x^2$.

The height of the triangle, from $(1,0)$ to point $B$, is $\tan x$, so comparing areas we get $x/2 \le (\tan x)/2 = \sin x / (2\cos x)$.

With a little algebra this becomes $\cos x \le (\sin x)/x$.

To cancel this common factor, we rationalize the denominator or numerator or both.

Example-3:-Evaluate Ans- Rationalizing the denominator and the numerator both = = = = 3. = If we are given a problem with we first of all note the highest st power of x in the whole question.

The condition says that $f(x)$ is trapped between $g(x)$ below and $h(x)$ above, and that at $x=a$, both $g$ and $h$ approach the same value.

This means the situation looks something like figure 4.3.1.

## Comments Solving Limit Problems

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