# Solving Limit Problems

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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With a little algebra this turns into $(\sin x)/x \le 1/\cos x$, giving us the $h$ we seek.

To find $g$, we note that the circular wedge is completely contained inside the larger triangle.

So now we have $$\cos x \le \le .$$ Finally, the two limits $\lim_\cos x$ and $\lim_1/\cos x$ are easy, because $\cos(0)=1$.

By the squeeze theorem, $\lim_ (\sin x)/x = 1$ as well.

This formula is not applicable when n is even ► Sometimes, we can directly use below formula to evaluate the limit ►If the degree of the numerator is more than or equal to the degree of the denominator, then we should divide.

Example-2:- Evaluate Ans:- = 10 Rationalization Method:- If we ever get 0/0 form in the problems involving square roots, then there must be a common factor in both numerator and denominator which must be canceled out to get a meaningful form.

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.

If f(x) and g(x) be two functions in such a way that (i) or (ii) both are continuous at x=a (iii) both are differentiable at x=a (iv) f'(x) and g'(x) are both continuous at x=a then If the result is still in 0/0 or in the form we can again differentiate and write like this You can download this PDF to get Practice Questions.

We want to compute this limit: $$\lim_ .$$ Equivalently, to make the notation a bit simpler, we can compute $$\lim_ .$$ In the original context we need to keep $x$ and $\Delta x$ separate, but here it doesn't hurt to rename $\Delta x$ to something more convenient.

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