Question

limit x-0(xcosx+sinx)/x^2+tanx​

Answer 1

Step-by-step explanation:

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Answer 2

The value of $\lim_{x \to 0} \dfrac{x\cos x+\sin x}{x^2+\​tan x}$ is 2.

Step-by-step explanation:

We have,

$\lim_{x \to 0} \dfrac{x\cos x+\sin x}{x^2+\​tan x}$

To find, the value of $\lim_{x \to 0} \dfrac{x\cos x+\sin x}{x^2+\​tan x}$ ?

∴ $\lim_{x \to 0} \dfrac{x\cos x+\sin x}{x^2+\​tan x}$ , $\dfrac{0}{0}$ form

Dividing numerator and denominator by x, we get

=$\lim_{x \to 0} \dfrac{\dfrac{x\cos x+\sin x}{x}}{\dfrac{x^2+\​tan x}{x}}$

$=\lim_{x \to 0} \dfrac{\cos x+\dfrac{\sin x}{x}}{x+\dfrac{\​tan x}{x}}$

We know that,

$\lim_{x \to 0} \dfrac{\sin x}{x}=1$ and

$\lim_{x \to 0} \dfrac{\tan x}{x}=1$

$=\lim_{x \to 0} \dfrac{\cos x+\dfrac{\sin x}{x}}{x+\dfrac{\​tan x}{x}}$

$=\lim_{x \to 0} \dfrac{\cos x+1}{x+1}$

Put x = 0, we get

= $\dfrac{\cos 0+1}{0+1}$

= $\dfrac{1+1}{1}$

= $\dfrac{2}{1}$

= 2

Thus, the value of $\lim_{x \to 0} \dfrac{x\cos x+\sin x}{x^2+\​tan x}$ is 2.