Consider the following :
$$\eqalign{
& 1.\,\mathop {\lim }\limits_{x \to 0} \frac{1}{x}{\text{ exists}}{\text{.}} \cr
& {\text{2}}{\text{.}}\,\mathop {\lim }\limits_{x \to 0} \,{e^{\frac{1}{x}}}{\text{ does not exist}}{\text{.}} \cr} $$
Which of the above is/are correct ?
A.
1 only
B.
2 only
C.
Both 1 and 2
D.
Neither 1 nor 2
Answer :
2 only
Solution :
$$\mathop {\lim }\limits_{x \to 0} \frac{1}{x} = \frac{1}{0} = \infty $$ which does not exist.
Hence, statements-1 is incorrect.
Now, $$\mathop {\lim }\limits_{x \to 0} \,{e^{\frac{1}{x}}} = {e^\infty }$$ which also does not exist.
Hence, statement-2 is correct.
Releted MCQ Question on Calculus >> Differentiability and Differentiation
Releted Question 1
There exist a function $$f\left( x \right),$$ satisfying $$f\left( 0 \right) = 1,\,f'\left( 0 \right) = - 1,\,f\left( x \right) > 0$$ for all $$x,$$ and-
A.
$$f''\left( x \right) > 0$$ for all $$x$$
B.
$$ - 1 < f''\left( x \right) < 0$$ for all $$x$$
C.
$$ - 2 \leqslant f''\left( x \right) \leqslant - 1$$ for all $$x$$
If $$f\left( a \right) = 2,\,f'\left( a \right) = 1,\,g\left( a \right) = - 1,\,g'\left( a \right) = 2,$$ then the value of $$\mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right)f\left( a \right) - g\left( a \right)f\left( x \right)}}{{x - a}}$$ is-
Let $$f:R \to R$$ be a differentiable function and $$f\left( 1 \right) = 4.$$ Then the value of $$\mathop {\lim }\limits_{x \to 1} \int\limits_4^{f\left( x \right)} {\frac{{2t}}{{x - 1}}} dt$$ is-