Question

You are given
$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}......$;
$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}......$;
$\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}......$
Then the value of
$\underset{x\to 0}{lim}\frac{x\cos x+\sin x}{x^2+\tan x}$ is

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Answer 1

Solution: We can do this problem using standard limits, which we will cover in the next topic. Here, we will show you how the standard expansions can be used to solve this problem.

We have,
$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}.....$
$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}......$
$\tan x=x+\frac{x^3}{3}+\frac{2x^5}{15}......$
We will substitute these values in our expression and simplify
$\Rightarrow \frac{x\cos x+\sin x}{x^2+\tan x}=\frac{x(1-\frac{x^2}{2!}+\frac{x^4}{4!}+..)+(x-\frac{x^3}{3!}+\frac{x^5}{5!}.....)}{x^2+(x+\frac{x^3}{3}+\frac{2x^5}{15}......)}$

$\Rightarrow \text{Then limit}=\underset{x\to 0}{lim}\frac{x\left[(1-\frac{x^2}{2!}+\frac{x^4}{4!}+..)+(1-\frac{x^2}{3!}+\frac{x^4}{5!}.....)\right]}{x(1+x+\frac{x^2}{3}+\frac{2x^4}{15}......)}$

$=\frac{1+1}{1}$

$=\frac{2}{1}=2$