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sachin nayak Grade: 12
`        proof of L'hospitals theorem`
7 years ago

## Answers : (2)

SAGAR SINGH - IIT DELHI
879 Points
```										Dear sachin,
If f and g are differentiable in a neighborhood of   x = c, and f(c) = g(c) = 0, then

provided the limit on the right exists. The same result holds for one-sided limits.
If f and g are differentiable and    f(x) =          g(x) =        - then

provided the last limit exists.
Proof:
The first part can be proved easily, if the right hand limit equals  f'(c) / g'(c): Since f(c) = g(c) = 0 we have

Taking the limit as x approaches c we get the first result. However,  the actual result is somewhat more general, and we have to be  slightly more careful. We will use a version of the Mean Value  theorem:
Take any sequence {xn}  converging to c from above. All assumptions of the generalized  Mean Value theorem are satisfied (check !) on [c, xn].  Therefore, for each n there exists a number cn in  the interval (c, xn)  such that

We are all  IITians and here to help you in your IIT  JEE preparation.  All the best.
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Sagar Singh
B.Tech IIT Delhi

```
7 years ago
Karthik Eyan
45 Points
```										Consider the linear approximation to f(x) and g(x) at x=a:

The ratio of these for x near a is:

which, if g'(a) is not 0 approaches f '(a) / g'(a) as x approaches a.
If g'(a) = 0 and f '(a) = 0 we can apply the same rule to the derivatives,    to give f "(a) / g"(a).
If these second derivatives are both 0 you can continue to higher derivatives,    etc. the result will be the ratio of the first pair of non-vanishing higher    derivatives at a.
Of course if the first non-vanishing derivative of the numerator is the kth    and occurs before the kth then the ratio is 0; if the first non-vanishing entry    of the denominator occurs after that of the numerator, the ratio goes to infinity    at a.
```
7 years ago
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