How to find $\lim_{x\to\infty}\frac{e^x+\cos x}{e^x-\sin x}$?

Question

$$\lim_{x\to\infty}\frac{e^x+\cos x}{e^x-\sin x}$$

Can someone help me find the limit without using the L'Hospital's rule?

Answer 1

$\displaystyle \lim_{x \to \infty} \dfrac{e^x + \cos x}{e^x - \sin x} = \lim_{x \to \infty} \dfrac{1 + e^{-x} \cos x}{1 - e^{-x}\sin x} = \dfrac{\lim_{x \to \infty}(1 + e^{-x} \cos x)}{\lim_{x \to \infty}(1 - e^{-x} \sin x)} = \dfrac{1}{1} = 1. \tag 1$

Answer 2

When you substitute $\infty$ in $e^x$ It becomes $\infty$ which is not desirable .

But we know that $\frac{1}{e^{\infty}}$ is equal to 0 .

So we divide $N^r$ And $D^r$ By $e^x$.

$$ \lim_{x\to \infty} \frac{1+\cos x\cdot e^{-x}}{1- \sin x \cdot e^{-x}} $$

Now ,

$$ lim_{x\to\infty} \frac{\cos x}{e^x} = \frac{-1<a\ number<1}{e^\infty}=0 $$

Similarly $$ lim_{x\to\infty} \frac{\sin x}{e^x} = \frac{-1<a\ number<1}{e^\infty}=0 $$

So your answer becomes

$$ lim_ {x\to\infty} \frac{1+0}{1-0}= 1 $$

And that's the final answer .

;-)

Answer 3

An option :

Let $x >0:$

$f(x)=\dfrac{e^x+\cos x}{e^x-\sin x}$.

$\dfrac{e^x-1}{e^x+1} \lt f(x) \lt \dfrac{e^x +1}{e^x-1}$.

$1- 2\dfrac{1}{e^x+1} \lt f(x) \lt 1+ 2\dfrac{1}{e^x-1}$

Take the limit $x \rightarrow \infty. $