Question

$\sf{Chapter:}$ Inverse Trigonometric Functions

$\sf{Class:}$ 12th​

Answer 1

hey there❤

Refer to attachment

Answer 2

Answer:

$\lim_{x \to 4} \dfrac{x^2-16}{\sqrt{x+5}-3}\\\\\implies \lim_{x \to 4} \dfrac{x^2-16}{\sqrt{x+5}-3}\times \dfrac{\sqrt{x+5}+3}{\sqrt{x+5}+3}\\\\\implies \lim_{x \to 4} \dfrac{(x+4)(x-4)(\sqrt{x+5}-3)}{(x+5-9)} \\\\\implies \lim_{x \to 4} (x+4)(\sqrt{x+5}+3)\\\\\implies (4+4)(\sqrt{4+5}+3)\\\\\implies 8 \times (3+3) \implies 8\times 6\\\\\implies 48$

Step-by-step explanation:

Firstly $lim_{x\to4}$ means that the value of x tends to be 4 . We can solve the above equation by putting x = 4 in the left hand side . If the limit had the indeterminate form of $\dfrac{x}{0}$ then we would have to simplify the denominator or the numerator so that we can get the limits value. Here we cannot put the value of x , as the limits is undefined and hence we can only found out the limiting value by simplification . The functional value can never be found for an undefined limits but the limits value can always be found .

Tell you what , limits is just cheating ( at least I feel so ) because you are stating x tends to 0 and not actually 0 and then you are putting 0 in the equation violating your statement XD .

Jokes apart , whenever we are given an indeterminate form of limits , we need to rationalize the undefined portion to get the limiting value of the function .