Sandwich Theorem: changing inequality

[AdityaDev]

Homework Statement

Using sandwich theorem evaluvate:
$$\lim_{x\rightarrow \infty} \frac{x+7sinx}{-2x+13}$$

Homework Equations

Sandwich theorem

The Attempt at a Solution

$-7 \leqslant 7sinx \leqslant 7$
$x-7 \leqslant x+7sinx \leqslant x+7$

Now my doubt: I want to divide the expression by $-2x+13$. But does this change the inequality? I don't know if it is positive or negative.
If I divide, I will get the expression in the question.
(hint given in my textbook: Inequality changes that is they assumed $-2x+13$ is negative.

Let me not change the inequality (I will get the answer but I need to know if what I am doing makes sense)

$\frac{x-7}{-2x+13} \leqslant \frac{x+7sinx}{-2x+13} \leqslant \frac{x+7}{-2x+13}$

$\lim_{x\rightarrow \infty} \frac{x-7}{-2x+13} = \lim_{x\rightarrow \infty} \frac{1- \frac {7}{x} }{-2+\frac{13}{x}} = -\frac{1}{2}$

Similarly the right side limit is also -1/2(same method). Hence limit of middle term is also -1/2. Answer is correct.

 

[Mark44]

Homework Statement

Using sandwich theorem evaluvate:
$$\lim_{x\rightarrow \infty} \frac{x+7sinx}{-2x+13}$$

Homework Equations

Sandwich theorem

The Attempt at a Solution

$-7 \leqslant 7sinx \leqslant 7$
$x-7 \leqslant x+7sinx \leqslant x+7$

Now my doubt: I want to divide the expression by $-2x+13$. But does this change the inequality? I don't know if it is positive or negative.
If I divide, I will get the expression in the question.
(hint given in my textbook: Inequality changes that is they assumed $-2x+13$ is negative.

Since your limit is as x gets very large, -2x + 13 can be assumed to be negative. Dividing by -2x + 13 will then change the direction of the inequalities.

Let me not change the inequality (I will get the answer but I need to know if what I am doing makes sense)

$\frac{x-7}{-2x+13} \leqslant \frac{x+7sinx}{-2x+13} \leqslant \frac{x+7}{-2x+13}$

$\lim_{x\rightarrow \infty} \frac{x-7}{-2x+13} = \lim_{x\rightarrow \infty} \frac{1- \frac {7}{x} }{-2+\frac{13}{x}} = -\frac{1}{2}$

Similarly the right side limit is also 1/2(same method). Hence limit of middle term is also 1/2. Answer is correct.