Why  is 0 and is 1?

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As h tends to 0+,-1/h tends to -infinity and e^-infinity tends to 0 and Hence the second limit is 1.Similarly,as h tends to 0-,-1/h tends to +infinity and so is e^-1/h.Hence the first limit is 0.

A similar concept was used in another problem posted by u.

hey man jab h0-to e ke power mein infinity aayega aur 1/infinity is 0

but jab h is 0+ toh e ke power mein -infinity aayega which means 1/(e)^infinity which is zero.

hope u got it.......