How to solve this limit without (L'Hôpital's rule)

Question

I need to solve this limit without L'Hôpital's rule. These questions always seem to have some algebraic trick which I just can't see this time. $$ \bbox[yellow] { \lim_{x\to 2} \left( \frac{e^2-e^x} {2-x} \right) } $$ Could someone give me a hint as to what I need to do to the fraction to make this work? Thanks!

Answer 1

The expression is the definition of the derivative of $e^x$ evaluated at $x=2$. It does not require L'Hopital to conclude that the limit is $e^2$.

Answer 2

since $\lim _{ x\to 0 } \frac { { e }^{ x }-1 }{ x } =1$ we have $$\lim _{ x\to 2 } \left( \frac { e^{ 2 }-e^{ x } }{ 2-x } \right) =\lim _{ x\to 2 } \frac { { e }^{ x }\left( { e }^{ 2-x }-1 \right) }{ 2-x } ={ e }^{ 2 }$$

Answer 3

Here are the steps $$\lim\limits_{x\to 2}\frac{e^2-e^x}{2-x}=e^2\lim\limits_{x\to 2}\frac{1-e^{x-2}}{2-x}$$ Let $h=x-2$, then $$e^2\lim\limits_{h\to 0}\frac{e^h-1}{h}=e^2$$

Answer 4

To find the result of this limit without redundancy or use other results you need start directly from the definition of $e^x$. There are several definitions. All of these equivalents. It remains to clarify which definition of $e^x$ you want to start calculating the limit in question. What definition of $e^x$ is acceptable to you? See the most common definitions here.

The most common definitions of $e^x$:

• $e^x=\left[\lim_{n\to \infty}\left(1+\frac{1}{n} \right)^{n}\right]^x, \quad n\in\mathbb{N}-\{0\}$ or $e^x=\left[\lim_{t\to \infty}\left(1+\frac{1}{t} \right)^{t}\right]^x, \quad t>0$,

• $e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\ldots \frac{x^k}{k!}+\ldots =\lim_{n\to \infty}\sum_{k=0}^n\frac{x^k}{k!}$

• inverse function of $\log (x)=\int_{0}^{x} \frac{1}{x}\mathrm{d}x$,

• unique function $y(x)$ such that $\frac{\mathrm{d} }{\mathrm{d}x}y(x)=y(x)$ and $y(0)=1$.

Suppose your definition of $e^t$ is $e^t=\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\ldots \frac{t^k}{k!}+\ldots =\lim_{n\to \infty}\sum_{k=0}^n\frac{t^k}{k!}$

Then $$ \lim_{x\to 2} \left( \frac{e^2-e^x}{2-x} \right) = \lim_{x\to 2} e^x\left( \frac{e^{2-x}-1}{2-x} \right) $$ By the property of limit for composition two functions (which I suppose is familiar to you) $$ \lim_{x\to 2} e^x\left( \frac{e^{2-x}-1}{2-x} \right) = \lim_{t\to 0} e^{t+2}\cdot\left( \frac{e^{t}-1}{t} \right) $$ Now note that $ \frac{e^{t}-1}{t}=1+\frac{t^1}{2!}+\frac{t^2}{3!}+\ldots \frac{t^{k-1}}{k!}+\ldots $ implies $ \lim_{t\to 0}\left(1+\frac{t^1}{2!}+\frac{t^2}{3!}+\ldots \frac{t^{k-1}}{k!}+\ldots \right)=1 $ and $ \lim_{t\to 0}\left( \frac{e^t-1}{t}\right)=1. $ By product property of two limits we have $$ \lim_{t\to 0} e^{t+2}\cdot\left( \frac{e^{t}-1}{t} \right) = \lim_{t\to 0} (e^{t+2})\cdot\lim_{t\to 0}\left( \frac{e^{t}-1}{t} \right) =e^{2}\cdot 1. $$

Answer 5

To make life simpler, define $x=2-y$ which makes $$A=\frac { e^{ 2 }-e^{ x } }{ 2-x }= \frac { e^{ 2 }-e^{ 2-y } }{y}=e^2\frac{1-e^{-y}}y$$ Now, use Taylor series around $y=0$ $$e^{-y}=1-y+\frac{y^2}{2}+O\left(y^3\right)$$ Replacing $$A=e^2\frac{1-\left(1-y+\frac{y^2}{2}+O\left(y^3\right) \right) } y=e^2\frac{y-\frac{y^2}{2}+O\left(y^3\right) } y=e^2\left(1-\frac{y}{2}+O\left(y^2\right)\right) $$ and, since $y\to 0$, you get the limit and also how it is approached.