Finding the existence of limit of a given function.

Question

I tried the L'Hospital Rule

Can any other decent method to solve this sum.

Answer 1

As Lord Shark has said, it is best that you actually expand $e^{\frac 1{x^3}}$ via its Taylor expansion.

Note that $e^{\frac 1{x^3}} = \sum_{n=0}^\infty \frac{1}{x^{3n}n!} = 1 + \frac 1{x^3} + \frac 1{2x^6} + O(x^{-9})$. Combining this with the rest of the expression tells you that: $$ x^n\left(e^{\frac 1{x^3}} - 1 + \frac 1{x^3}\right) = 2x^{n-3} + \frac 12x^{n-6} + O(x^{n-9}) $$

If $n < 3$, then the limit most definitely exists, since the expression above is at least $O(x^{-1})$.

Even when $n=3$, the limit will be $2$, since the rest is $O(x^{-3})$.

However, when $n>3$, the first term will certainly go to $\infty$, and the rest are just positive terms, so the limit will not exist.

It follows that all the above options are true. (In competitive exams, I think this approach works better than L'Hopital, in case this is a competitive exam).

Answer 2

It is from IIT-JAM (entrance exam in india for masters)

we have to find limit of $f(x)$ at $\infty$

Part (a)

as we can see $f(x)$ is of form of $\infty \times 0$ we have to write it in the form of $0 \over 0$ to use L'Hopitals Rule so we can write the function as

$$f(x)= {e^{1\over x^3}- 1+ {1\over x^3}\over {1\over x}}$$

Now using L'Hopitals Rule i.e we are going to differentiate both numerator and denominator with respect to $x$, we get

$${-3\over x^4} \bigg( {e^{1\over x^3}} +1 \bigg) \over {-1\over x^2} $$

by further simplifying it we get

$${3\over x^2} \bigg( {e^{1\over x^3}} +1 \bigg) $$ so now we have

$$\lim_{x\to \infty}f(x)= \lim_{x\to \infty} {3\over x^2} \bigg( {e^{1\over x^3}} +1 \bigg)=0 $$

Part (b)

$$xf(x)= {e^{1\over x^3}- 1+ {1\over x^3}\over {1\over x^2}} $$ so again we have $0\over 0 $ form, so differentiating Numerator and denominator with respect to x we get

$$=\lim_{x\to \infty}{6\over x} \bigg( {e^{1\over x^3}} +1 \bigg)=0$$

Part (c)

$$\lim_{x\to \infty}x^2f(x)= \lim_{x\to \infty} f(x)= {e^{1\over x^3}- 1+ {1\over x^3}\over {1\over x^3}}$$

so again we have $0\over 0 $ form, so differentiating Numerator and denominator with respect to x we get

$$\lim_{x\to \infty} ({e^{1\over x^3}} +1) =2 $$

Part (d)

but for $x$ to the power m where $m>2$ ,limit doesnot exist because then x will dominate lets take $x=m>2$ for example

$$\lim_{x\to \infty}x^mf(x)= \lim_{x\to \infty} x^{m-2}{3\over m+1} \bigg( {e^{1\over x^3}} +1 \bigg)=\infty (for \quad m>2) $$ so our all four options $(a),(b),(c)$ and $(d)$ are correct

we can do the Part (d) first interpret others easily because we see in the last part for m-2>0 limit doesnot exist i.e for m>2 limit doesnot exist