Download App

STD 11 - 12. limits — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 12. limits4 Marks
MCQ
The value of $\mathop {\lim }\limits_{x \to a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$ is
  • ✓
    $1$
  • B
    $-1$
  • C
    $0$
  • D
    None of these
✓

Answer

Correct option: A.
$1$
a
(a) $\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to a} \,\,\frac{{\log \,(x - a)}}{{\log \,({e^x} - {e^a})}} = \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to a} \,\,\frac{{{e^x} - {e^a}}}{{(x - a)\,{e^x}}}$, $\left( {{\rm{Form}} \,\, \frac{0}{0}} \right)$

$ = \mathop {\lim }\limits_{x \to a} \,\,\frac{{{e^x}}}{{\left\{ {(x - a)\,{e^x} + {e^x}} \right\}}} = \frac{{{e^a}}}{{{e^a}}} = 1.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The coefficient of $x^{91}$ in the series $^{100}{C_1}\,{2^8}.\,{\left( {1\, - \,x} \right)^{99}}\, + {\,^{100}}{C_2}\,{2^7}.\,{\left( {1\, - \,x} \right)^{98}}\, + {\,^{100}}{C_3}\,{2^6}.\,{\left( {1\, - \,x} \right)^{97}}\, + \,....\, + {\,^{100}}{C_9}\,{\left( {1\, - \,x} \right)^{91}}$ is equal to -
→
The centre of the ellipse$\frac{{{{(x + y - 2)}^2}}}{9} + \frac{{{{(x - y)}^2}}}{{16}} = 1$ is
→
$\int {\,\,\frac{{{{\cot }^{ - 1}}({e^x})}}{{{e^x}}}} $ $dx $ is equal to :
→
The sum of the terms of an infinitely decreasing geometric progression is equal to the greatest value of the function $f (x) = x^3 + 3x -9$ on the interval $[- 2, 3]$ . If the difference between the first and the second term of the progression is equal to $f ' (0)$ then the common ratio of the $G.P$. is
→
Let $f(x)=\left\{\begin{array}{ll}x^2\left|\cos \frac{\pi}{x}\right|, & x \neq 0 \\ 0, & x=0\end{array}, x \in I R\right.$, then $f$ is
→
The range of the function $f(x) = \frac{{\sqrt {1 - {x^2}} }}{{1 + \left| x \right|}}$ is 
→
If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $I_1=\int_a^b x \sin \left(4 x-x^2\right) d x, I_2=\int_a^b \sin \left(4 x-x^2\right) d x$ , then $36 \frac{I_1}{I_2}$ is equal to :
→
The area of the triangle whose vertices are represented by the complex numbers $0, z,$ $z{e^{i\alpha }},$ $(0 < \alpha < \pi )$ equals
→
The least positive integer $\mathrm{n}$ such that $\frac{(2 \mathrm{i})^{\mathrm{n}}}{(1-\mathrm{i})^{\mathrm{n}-2}}, \mathrm{i}=\sqrt{-1}$ is a positive integer, is ..... .
→
If $\mathrm{y}=\mathrm{y}(\mathrm{x})$ is the solution of the differential equation, $\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-1\right)=\mathrm{e}^{\mathrm{x}}$ such that $\mathrm{y}(0)=0,$ then $\mathrm{y}(1)$ is equal to 
→