test series

Question

The answer given is 1.5 (they are using L'Hospital Rule)

But mine is 1 [ (e^x-1)/x=1   And Log(1+x)/x = 1]

Both looks but the answer is different please verify which one is correct. ASAP

Answer 1

Put $x = 0$ to get $\dfrac{0\cdot e^0 - log(1+0)}{0^2}$ which is a $\frac00$ form, so L'Hôpital's (pronounced Lopital) rule can be applied.

$$\begin{align*}
&\lim\limits_{x \to 0} \dfrac{xe^x - \log(1+x)}{x^2} \\[1.5em]
=& \lim_{x \to 0} \frac{\frac{d}{dx}(xe^x - \log(1+x))}{\frac{d}{dx}x^2} & \{ \text{L'Hôpital's rule} \\[1.5em]
=& \lim_{x \to 0} \frac{(e^x + xe^x) - \frac{1}{1+x}}{2x}
\end{align*}$$
 

Put $x=0$ to get $\dfrac{(e^0+0\cdot e^0) - \frac1{1+0}}{2\cdot 0}$ which is again of $\frac00$ form, so we need to apply L'Hôpital's rule again.

$$\begin{align*}
=& \lim_{x \to 0} \frac{\frac{d}{dx} \left((e^x + xe^x) - \frac{1}{1+x}\right)}{\frac{d}{dx}(2x)} & \{ \text{L'Hôpital's rule} \\[1.5em]
=& \lim_{x \to 0} \frac{\Bigl(e^x+(e^x+xe^x)\Bigr) - \left(-1\cdot \frac1{(1+x)^2}\right)}{2}\\
\end{align*}$$

 

Put $x=0$ to get $\dfrac{(e^0+(e^0+0\cdot e^0)) + \frac1{(1+0)^2}}{2}$ which is $\frac{(1+(1)) + 1}{2} = \frac32 = 1.50$ (two decimal places)

 

Edit: There's an important lesson in the comments, so make sure you read that.

Comments

https://www.google.com/search?q=%28x*e%5Ex-ln%281%2Bx%29%29%2F%28x%5E2%29

 

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what is the problem with this approach? @Pragy Agarwal

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Ah, you're making the subtle but classic mistake of incorrectly using the Linearity of Limits.

Linearity of Limits: The limit of the sum is the sum of the limits, provided the limits exist.

That is, if both $\lim\limits_{x \to c} f(x)$ and $\lim\limits_{x \to c} g(x)$ exist, then and only then, $$\lim_{x\to c} \Bigl [ f(x) + g(x) \Bigr ] = \lim_{x \to c} f(x) + \lim_{x \to c} g(x)$$

 

In this question, we have a sum of limits $$\lim_{x \to 0} \left [ \frac{xe^x}{x^2} - \frac{\log(1+x)}{x^2} \right ]$$

However, the neither of the individual limits exist! $$\lim_{x \to 0} \dfrac{xe^x}{x^2} = \lim_{x \to 0} \dfrac{e^x}{x} = \dfrac{1}{0} =  \infty$$ and  $$\lim_{x \to 0} \frac{\log{(1+x)}}{x^2} = \lim_{x \to 0} \frac{-\frac{1}{1+x}}{2x} = -\frac10 = -\infty$$

 

Since the individual limits don't exist, we can't use the Linearity of Limits!

 

So your step 3 of splitting the limits, and your step 5 of combining the limits back are both wrong.

 

Also, in step 3, you split $\left (\lim\limits_{x \to 0} \dfrac{\log(1+x)}{x^2} \right)$ to get $\left (\dfrac1x \cdot \lim\limits_{x\to 0} \dfrac {\log(1+x)}{x} \right)$ which is also incorrect, since you can't simply pull the $\dfrac1x$ out of the limit.

 

Morals of the story:

1. Linearity of limits can only be applied when the individual limits exist

2. Don't take shortcuts when doing math. Properly write out each term. You chose to omit putting the limits besides each term, and ended up incorrectly pulling out a $\dfrac1x$

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Thank you very much for pointing out the mistake and clarifying the doubt.