GATE CSE 2017 | Set 1 | Question: 28

Question

The value of $\displaystyle \lim_{x\rightarrow 1} \frac{x^{7}-2x^{5}+1}{x^{3}-3x^{2}+2}$

• A. is $0$
• B. is $-1$
• C. is $1$
• D. does not exist

Comments

This question is the main Concept L-H Rule.

0/0 form

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For this question, how to check whether the limit exists or not?
How to perform LHL=RHL? I know we find the limit to be 1 but since we have an option which says limit may not exist, don't we need to check LHL=RHL or not?

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I think ,  if there is an option of Limit Does Not Exist (DNE) , then definitely we should  have to check whether limit exists or not.

Here, Left Hand Limit (LHL) = $\lim_{x\rightarrow 1^{-}}\frac{x^{7}-2x^{5} + 1}{x^{3}-3x^{2}+2} = \lim_{h\rightarrow 0}\frac{(1-h)^{7}-2(1-h)^{5} + 1}{(1-h)^{3}-3(1-h)^{2}+2}$

 $= \lim_{h\rightarrow 0}\frac{-7(1-h)^{6}+10(1-h)^{4} }{-3(1-h)^{2}+6(1-h)}$ (Using L'H$\hat{o}$pital's Rule)

= $\frac{3}{3}$ = 1

Similarly , Right Hand Limit(RHL) =

$\lim_{x\rightarrow 1^{+}}\frac{x^{7}-2x^{5} + 1}{x^{3}-3x^{2}+2} = \lim_{h\rightarrow 0}\frac{(1+h)^{7}-2(1+h)^{5} + 1}{(1+h)^{3}-3(1+h)^{2}+2}$

 $= \lim_{h\rightarrow 0}\frac{7(1+h)^{6}-10(1+h)^{4} }{3(1+h)^{2}-6(1+h)}$ (Using L'H$\hat{o}$pital's Rule)

= $\frac{-3}{-3}$ = 1

So, here LHL = RHL . It means Limit exists.

Answer 1

Since substituting $x=1$ we get $\frac{0}{0}$ which is indeterminate.

After applying L'Hospital rule, we get $\dfrac{(7x^{6} -10x{^4})}{(3x^{2} - 6x)}$

Now substituting $x=1$ we get $\left(\frac{-3}{-3}\right) =1.$

Hence, answer is $1$.

Correct Answer: $C$

Answer 2

correct option is c

Comments

Nice handwriting, wow beautiful explanation

Answer 3

solution...........

Answer 4

Correct Answer is 1 Put the limit and check whether..... it 0/0 or not 

after checking apply l'hospitals  onit