Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Without using the method of L’ Hospital, find the limit <img alt="\lim _{x \rightarrow 0} \frac{a^{1-\sin x}-a^{1-\tan x}}{e^{1-\sin x}-e^{1-\tan x}}" src="https

Without using the method of L’ Hospital, find the limit

\lim _{x \rightarrow 0} \frac{a^{1-\sin x}-a^{1-\tan x}}{e^{1-\sin x}-e^{1-\tan x}}

Option: 1

\ln a^{\frac{a}{e}}


Option: 2

a^4


Option: 3

\ln a^a


Option: 4

\frac{1}{\ln a^a}


Answers (1)

best_answer

Note the following exponential limits.

\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^x-1}{x}=1 \\ & \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log _c a=\ln a \end{aligned}

Now, evaluate the following limit using the above formula.

\begin{aligned} & =\frac{\lim _{x \rightarrow 0} \frac{a^{1 \tan x}\left(a^{\tan x \sin x}-1\right)}{\tan x-\sin x}}{\lim _{x \rightarrow 0} \frac{e^{1 \tan x}\left(e^{\tan x \sin x}-1\right)}{\tan x-\sin x}} \\ & =\frac{\lim _{x \rightarrow 0} a^{1-\tan x} \times \lim _{x \rightarrow 0} \frac{a^{\tan x-\sin x}-1}{\tan x-\sin x}}{\lim _{x \rightarrow 0} e^{1-\tan x} \times \lim _{x \rightarrow 0} \frac{e^{\tan x-\sin x}-1}{\tan x-\sin x}} \\ & =\frac{a^{1-0} \times \ln a}{e^{1-0} \times 1} \quad[x \rightarrow 0 \Rightarrow \tan x \rightarrow 0, \sin x \rightarrow 0] \\ & =\frac{a}{e} \ln a \\ & =\ln a^{\frac{a}{e}} \\ & \end{aligned}

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 final answer key likely in February 2nd week; result date, marking scheme
anam-khan

JEE Mains 2025 Answer key (OUT) LIVE: Session 1 result by February 12 on jeemain.nta.nic.in; cut-off
anam-khan

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?

Latest Question