Why do i get the wrong answer if i evaluate the following limit in the following way -

Question

The actual answer of this is e^-8 .But why do i get the answer as 1 when i evaluate the following limit like this -

Answer 1

Your proof is wrong in going from line 2 to line 3. You say you are just using the "algebra of limits" there, but that is not true.

The algebra of limits says if $f,g$ have limits $L,M$ at $a,$ then $\lim_{x\to a}(f(x)+g(x)) = L+M,$ $\lim_{x\to a}(f(x)g(x)) =LM,$ etc. Which of these basic results allows us to go from line 2 to line 3?

Here's something to consider: Line 2, after some simplifying, is

$$\tag 1 \left (\frac{(1+u)^{1/u}}{e}\right )^{16/\sin u}.$$

As you may recall, $(1+u)^{1/u}/e<1$ for small positive $u.$ And $16/\sin u$ is huge for small positive $u.$ So we're taking something in $(0,1)$ and raising it to a very large power. It's not surprising that the answer turns out to be a number less than $1.$

On to the desired limit: It's good to recall logarithms are wonderful things in this kind of context. So apply $\ln$ to $(1)$ to get

$$\frac{16}{\sin u}\left (\frac{1}{u}\ln (1+u)-1\right ).$$

Now Taylor tells us $\ln(1+u)=u-u^2/2 +O(u^3)$ as $u\to 0.$ A little bit of work then shows the above equals

$$\frac{1}{\sin u}\left (-8u + O(u^2)\right )=-8\frac{u}{\sin u} + \frac{O(u^2)}{\sin u}.$$

The limit of this as $u\to 0$ is $-8\cdot 1+0=-8.$ Exponentiating back gives $e^{-8}$ for the desired limit.

Answer 2

When you take a limit of an expression, you can't take the limit of just a part of the expression (such as the numerator) and leave the rest of the expression the same. That seems to be what you did in the third line.

To see why this is wrong, just consider this example: $$\lim_{x \to 0}\frac{x}{x}$$ If I just take the limit in the numerator only, I get $$\lim_{x\to 0} \frac{0}{x}$$ Obviously these are not equal - the first one is 1 and the second is 0.

Answer 3

You made a mistake when you wrote that

$$(1+x)^{\frac{16}{x\sin(x)}}=$$

$$e^{\frac{16}{x\sin(x)}\ln(1+x)}\sim e^{\frac{16}{\sin(x)}}$$

because $\ln(1+x)\sim x$.

In fact we have

$$e^{f(x)}\sim e^{g(x)}\iff \lim (f(x)-g(x))=0$$

Answer 4

The expression whose limit you want to calculate is:

$\lim_{n\to 0}\Big(\frac{(1+n)^{2/n}}{e^{2}}\Big)^{\frac{8}{sin(n)}}$.

$\lim_{n\to 0}\Big(e^{\frac{-1}{sin(n)}}(n+1)^{\frac{1}{n.sin(n)}}\Big)^{16}$,

$=(e^{-1/2)})^{16}=e^{-8}$