Question

Solve this question urgent

Answer 1

Answer:

$\mathsf{\lim_{x \to 0}\dfrac{tanx}{x}}\\\\\implies \mathsf{\lim_{x \to 0}\dfrac{\dfrac{sinx}{cosx}}{x}}\\\\\implies \mathsf{\lim_{x \to 0}\dfrac{sinx}{xcosx}}\\\\\implies \mathsf{\lim_{x \to 0}\dfrac{sinx}{x}\times\mathsf{\lim_{x \to 0}}\dfrac{1}{cosx}}\\\\\implies 1\times 1 \\\\\implies 1$

Step-by-step explanation:

Use the basic formula of trigonometry to start with .

Use tan A = sin A / cos A .

Note that sin x / x tends to be 1 when x tends to be 0 .

Also note that :

cos x tends to be 1 when x tends to be 0 .

Hence we use the above formulas to get the limits solved .

Final answer is 1 .

Answer 2

heya..

Use the basic formula of trigonometry to start with .

Use tan A = sin A / cos A .

Note that sin x / x tends to be 1 when x tends to be 0 .

Also note that :

cos x tends to be 1 when x tends to be 0 .

Hence we use the above formulas to get the limits solved .

Final answer is 1 ..