/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 250 $$ \lim _{x \rightarrow 0} \fr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 250

$$ \lim _{x \rightarrow 0} \frac{1}{\sin x}-\frac{1}{\tan x}\\{\text { Ans. } 0\\} $$

Short Answer

Expert verified
To find the limit of \(\lim _{x \rightarrow 0} \frac{1}{\sin x}-\frac{1}{\tan x}\), we rewrite the function using a common denominator, simplify the expression, and apply L'Hopital's Rule. After these steps, we find that the limit is 0: \(\lim _{x \rightarrow 0} \frac{1}{\sin x} - \frac{1}{\tan x}=0\).

Step by step solution

01

Rewrite the function using a common denominator

We can rewrite the function as a single fraction with a common denominator. The common denominator for sin(x) and tan(x) is sin(x)tan(x): \(\lim _{x \rightarrow 0} \frac{1}{\sin x}-\frac{1}{\tan x} = \lim _{x \rightarrow 0} \frac{\tan(x)-\sin(x)}{\sin(x) \tan(x)}\)
02

Simplify the expression

Now, we can simplify the expression, replacing tan(x) with sin(x)/cos(x): \(\lim _{x \rightarrow 0} \frac{\tan(x)-\sin(x)}{\sin(x) \tan(x)} = \lim_{x \rightarrow 0} \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{\sin(x) \cdot \frac{\sin(x)}{\cos(x)}}\)
03

Cancel terms and further simplify

Here, we can cancel the sin(x) terms from both the numerator and denominator: \(\lim_{x \rightarrow 0} \frac{\frac{\sin(x)}{\cos(x)} - \sin(x)}{\sin^{2}(x) / \cos(x)} = \lim_{x \rightarrow 0} \frac{\cos(x) - 1}{\sin^{2}(x)}\)
04

Apply L'Hopital's Rule

Now, we can apply L'Hopital's Rule as we have an indeterminate form of 0/0. We take the derivative of both the numerator and denominator until we get an expression without the indeterminate form: \(\lim_{x \rightarrow 0} \frac{-\sin(x)}{2 \sin(x) \cos(x)}\) As x approaches 0, sin(x) approaches 0. Then the limit becomes: \(\lim_{x \rightarrow 0} \frac{-\sin(x)}{2 \sin(x) \cos(x)} = \lim_{x \rightarrow 0} \frac{-1}{2\cos(x)}\) The limit of this expression is 0 as the sine function approaches 0, and the cosine function approaches 1. So, the final answer is:
05

State the result

\(\lim _{x \rightarrow 0} \frac{1}{\sin x} - \frac{1}{\tan x} = 0\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

$$ \lim _{x \rightarrow 1} \ln (\sin \pi x-\cos \pi x) \quad\\{\text { Ans. } 0\\} $$

$$ \lim _{x \rightarrow 0}\left(\cos ^{-1} x\right)^{\cos x}\left\\{\text { Ans. } \frac{\pi}{2}\right\\} $$

$$ \lim _{x \rightarrow \infty} x+e^{x} \quad\\{\text { Ans. } \infty\\} $$

$$ \left.\lim _{x \rightarrow 0} \frac{2^{x}+\sin x}{\cos ^{-1} x+\tan ^{-1} x} \text { \\{Ans. } \frac{2}{\pi}\right\\} $$

$$ \lim _{x \rightarrow 1} \frac{1}{1-x}-\frac{3}{1-x^{3}}\\{\text { Ans. }-1\\} $$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.